answer:<think>Alright, so I have this problem about a technology coordinator integrating robotics into a high school curriculum. There are two parts to this problem. Let me tackle them one by one.Starting with the first part: The school has 15 classes, each needing a different number of robots. The number of robots required by each class follows a quadratic sequence defined by ( R_n = an^2 + bn + c ). The first three classes require 3, 7, and 13 robots respectively. I need to find the values of ( a ), ( b ), and ( c ).Okay, so since it's a quadratic sequence, each term is given by a quadratic formula. The first three terms are given, so I can set up equations based on those.Let me write down the equations for n=1, n=2, and n=3.For n=1:( R_1 = a(1)^2 + b(1) + c = a + b + c = 3 )For n=2:( R_2 = a(2)^2 + b(2) + c = 4a + 2b + c = 7 )For n=3:( R_3 = a(3)^2 + b(3) + c = 9a + 3b + c = 13 )So now I have a system of three equations:1. ( a + b + c = 3 )2. ( 4a + 2b + c = 7 )3. ( 9a + 3b + c = 13 )I need to solve this system for a, b, and c.Let me subtract equation 1 from equation 2 to eliminate c:Equation 2 - Equation 1:( (4a + 2b + c) - (a + b + c) = 7 - 3 )Simplify:( 3a + b = 4 ) ... Let's call this equation 4.Similarly, subtract equation 2 from equation 3:Equation 3 - Equation 2:( (9a + 3b + c) - (4a + 2b + c) = 13 - 7 )Simplify:( 5a + b = 6 ) ... Let's call this equation 5.Now we have two equations:4. ( 3a + b = 4 )5. ( 5a + b = 6 )Subtract equation 4 from equation 5:( (5a + b) - (3a + b) = 6 - 4 )Simplify:( 2a = 2 )So, ( a = 1 )Now plug a=1 into equation 4:( 3(1) + b = 4 )( 3 + b = 4 )So, ( b = 1 )Now plug a=1 and b=1 into equation 1:( 1 + 1 + c = 3 )( 2 + c = 3 )So, ( c = 1 )Therefore, the quadratic formula is ( R_n = n^2 + n + 1 ).Let me verify this with the given terms:For n=1: 1 + 1 + 1 = 3 ✔️For n=2: 4 + 2 + 1 = 7 ✔️For n=3: 9 + 3 + 1 = 13 ✔️Looks good. So, a=1, b=1, c=1.Moving on to the second part: The coordinator plans to set up a centralized control system that requires each robot to synchronize every 5 minutes for a software update. The synchronization process takes t minutes and follows a logistic growth model given by ( P(t) = frac{L}{1 + e^{-k(t - t_0)}} ). We are given that the system can handle 50 robots at t=10 minutes and 90 robots at t=20 minutes. We need to find L, k, and t0.So, the logistic growth model is given, and we have two points: (10,50) and (20,90). We need to solve for L, k, and t0.First, let's recall that the logistic function has an inflection point at t0, where the growth rate is maximum. At t0, the function is equal to L/2. So, if we can find t0, that would be helpful.But we don't know t0 yet. So, let's set up the equations.Given:1. At t=10, P=50:( 50 = frac{L}{1 + e^{-k(10 - t_0)}} )2. At t=20, P=90:( 90 = frac{L}{1 + e^{-k(20 - t_0)}} )We have two equations with three unknowns: L, k, t0. So, we need another equation or a way to relate them.But wait, perhaps we can express both equations in terms of L and then take a ratio to eliminate L.Let me denote equation 1 as:( 50 = frac{L}{1 + e^{-k(10 - t_0)}} ) --> Let's call this equation A.Equation 2 as:( 90 = frac{L}{1 + e^{-k(20 - t_0)}} ) --> Let's call this equation B.Let me solve equation A for L:From A: ( L = 50 left(1 + e^{-k(10 - t_0)}right) )Similarly, from B: ( L = 90 left(1 + e^{-k(20 - t_0)}right) )Since both equal L, set them equal:( 50 left(1 + e^{-k(10 - t_0)}right) = 90 left(1 + e^{-k(20 - t_0)}right) )Let me divide both sides by 10 to simplify:( 5 left(1 + e^{-k(10 - t_0)}right) = 9 left(1 + e^{-k(20 - t_0)}right) )Let me denote ( x = e^{-k(10 - t_0)} ). Then, note that ( e^{-k(20 - t_0)} = e^{-k(10 - t_0)} cdot e^{-k(10)} = x cdot e^{-10k} ).So, substituting x into the equation:( 5(1 + x) = 9(1 + x e^{-10k}) )Let me expand both sides:Left side: 5 + 5xRight side: 9 + 9x e^{-10k}Bring all terms to the left:5 + 5x - 9 - 9x e^{-10k} = 0Simplify:-4 + 5x - 9x e^{-10k} = 0Factor out x:-4 + x(5 - 9 e^{-10k}) = 0So,x(5 - 9 e^{-10k}) = 4But x = e^{-k(10 - t0)}, so:e^{-k(10 - t0)} (5 - 9 e^{-10k}) = 4Hmm, this seems a bit complicated. Maybe I can express t0 in terms of k or something else.Wait, another approach: Let's take the ratio of equation A and equation B.From A: ( 50 = frac{L}{1 + e^{-k(10 - t_0)}} )From B: ( 90 = frac{L}{1 + e^{-k(20 - t_0)}} )Divide A by B:( frac{50}{90} = frac{1 + e^{-k(20 - t_0)}}{1 + e^{-k(10 - t_0)}} )Simplify:( frac{5}{9} = frac{1 + e^{-k(20 - t_0)}}{1 + e^{-k(10 - t_0)}} )Let me denote ( y = e^{-k(10 - t_0)} ). Then, ( e^{-k(20 - t_0)} = e^{-k(10 - t_0)} cdot e^{-10k} = y e^{-10k} ).So, substituting y:( frac{5}{9} = frac{1 + y e^{-10k}}{1 + y} )Cross-multiplied:5(1 + y) = 9(1 + y e^{-10k})Expand:5 + 5y = 9 + 9y e^{-10k}Bring all terms to left:5 + 5y - 9 - 9y e^{-10k} = 0Simplify:-4 + 5y - 9y e^{-10k} = 0Factor y:-4 + y(5 - 9 e^{-10k}) = 0So,y(5 - 9 e^{-10k}) = 4But y = e^{-k(10 - t0)}, so:e^{-k(10 - t0)} (5 - 9 e^{-10k}) = 4This seems similar to what I had earlier. Hmm.Alternatively, maybe I can express t0 in terms of k.Let me think: The logistic function has its midpoint at t0, so when t = t0, P(t0) = L/2.But we don't have a data point at t0. So, maybe we can assume that t0 is somewhere between 10 and 20, but we don't know.Alternatively, let's take the two equations:From A: ( 50 = frac{L}{1 + e^{-k(10 - t_0)}} )From B: ( 90 = frac{L}{1 + e^{-k(20 - t_0)}} )Let me solve for e^{-k(10 - t0)} and e^{-k(20 - t0)}.From A:( 1 + e^{-k(10 - t0)} = frac{L}{50} )So,( e^{-k(10 - t0)} = frac{L}{50} - 1 )Similarly, from B:( 1 + e^{-k(20 - t0)} = frac{L}{90} )So,( e^{-k(20 - t0)} = frac{L}{90} - 1 )Now, note that ( e^{-k(20 - t0)} = e^{-k(10 - t0)} cdot e^{-10k} )So,( left( frac{L}{50} - 1 right) e^{-10k} = frac{L}{90} - 1 )Let me denote ( frac{L}{50} - 1 = A ) and ( frac{L}{90} - 1 = B ), so:( A e^{-10k} = B )But A and B are related to L. Let me express A and B in terms of L:A = (L/50) - 1B = (L/90) - 1So,[(L/50) - 1] e^{-10k} = (L/90) - 1Let me write this as:[(L - 50)/50] e^{-10k} = (L - 90)/90Multiply both sides by 50*90 to eliminate denominators:(L - 50)*90 e^{-10k} = (L - 90)*50Simplify:90(L - 50) e^{-10k} = 50(L - 90)Divide both sides by 10:9(L - 50) e^{-10k} = 5(L - 90)Let me expand both sides:9L - 450) e^{-10k} = 5L - 450Bring all terms to left:9L e^{-10k} - 450 e^{-10k} - 5L + 450 = 0Factor terms:(9L e^{-10k} - 5L) + (-450 e^{-10k} + 450) = 0Factor L and 450:L(9 e^{-10k} - 5) + 450(-e^{-10k} + 1) = 0Let me factor out the negative sign in the second term:L(9 e^{-10k} - 5) - 450(e^{-10k} - 1) = 0Hmm, this seems a bit messy. Maybe I can express e^{-10k} as a variable, say, let’s let z = e^{-10k}Then, the equation becomes:L(9z - 5) - 450(z - 1) = 0So,9L z - 5L - 450z + 450 = 0Let me rearrange terms:(9L z - 450z) + (-5L + 450) = 0Factor z from first two terms:z(9L - 450) + (-5L + 450) = 0Factor 9 from the first term:z(9(L - 50)) + (-5L + 450) = 0So,9(L - 50) z + (-5L + 450) = 0Let me write this as:9(L - 50) z = 5L - 450Divide both sides by (L - 50):9 z = (5L - 450)/(L - 50)Simplify numerator:5L - 450 = 5(L - 90)Denominator: L - 50So,9 z = 5(L - 90)/(L - 50)But z = e^{-10k}, so:9 e^{-10k} = 5(L - 90)/(L - 50)Let me solve for e^{-10k}:e^{-10k} = [5(L - 90)] / [9(L - 50)]Now, recall that from equation A:From A: ( 50 = frac{L}{1 + e^{-k(10 - t0)}} )So,( 1 + e^{-k(10 - t0)} = L / 50 )Thus,( e^{-k(10 - t0)} = (L / 50) - 1 )Similarly, from equation B:( e^{-k(20 - t0)} = (L / 90) - 1 )But also, ( e^{-k(20 - t0)} = e^{-k(10 - t0)} cdot e^{-10k} )So,( (L / 90 - 1) = (L / 50 - 1) cdot e^{-10k} )But we have an expression for e^{-10k} from earlier:e^{-10k} = [5(L - 90)] / [9(L - 50)]So,( (L / 90 - 1) = (L / 50 - 1) cdot [5(L - 90) / 9(L - 50)] )Let me compute both sides.Left side: L/90 - 1 = (L - 90)/90Right side: (L/50 - 1) * [5(L - 90)/9(L - 50)]Compute (L/50 - 1) = (L - 50)/50So,Right side: [(L - 50)/50] * [5(L - 90)/9(L - 50)] = [ (L - 50) * 5(L - 90) ] / [50 * 9(L - 50) ]Simplify:The (L - 50) terms cancel out, and 5/50 = 1/10:= [5(L - 90)] / [9 * 10] = (L - 90)/18So, right side is (L - 90)/18Thus, equation becomes:(L - 90)/90 = (L - 90)/18Multiply both sides by 90 to eliminate denominators:(L - 90) = 5(L - 90)Wait, that can't be right. Let me check my steps.Wait, left side: (L - 90)/90Right side: (L - 90)/18So,(L - 90)/90 = (L - 90)/18Multiply both sides by 90*18 to eliminate denominators:18(L - 90) = 90(L - 90)Simplify:18L - 1620 = 90L - 8100Bring all terms to left:18L - 1620 - 90L + 8100 = 0Combine like terms:-72L + 6480 = 0So,-72L = -6480Divide both sides by -72:L = (-6480)/(-72) = 6480 / 72Calculate:6480 ÷ 72: 72*90=6480, so L=90.Wait, L=90? But in the logistic function, L is the maximum capacity. But we have P(t)=90 at t=20. So, if L=90, that would mean the maximum capacity is 90, which is achieved at t=20. But in the logistic model, the function approaches L asymptotically, but never exceeds it. So, if at t=20, P(t)=90, which is equal to L, that would mean that at t=20, the system is at maximum capacity.But let's check if this makes sense.If L=90, then let's go back to equation A:50 = 90 / (1 + e^{-k(10 - t0)})So,1 + e^{-k(10 - t0)} = 90 / 50 = 1.8Thus,e^{-k(10 - t0)} = 0.8Similarly, from equation B:90 = 90 / (1 + e^{-k(20 - t0)})So,1 + e^{-k(20 - t0)} = 1Thus,e^{-k(20 - t0)} = 0But e^{-k(20 - t0)} = 0 implies that -k(20 - t0) approaches negative infinity, which would require k to be positive and (20 - t0) to approach positive infinity, which isn't practical. Alternatively, it implies that t0 = 20, because if t0=20, then e^{-k(20 - 20)}=e^0=1, but that would make 1 + 1=2, so 90=90/2=45, which contradicts. Wait, no, if t0=20, then equation B would be:90 = 90 / (1 + e^{-k(20 - 20)}) = 90 / (1 + 1) = 45, which is not 90. So, that can't be.Wait, perhaps I made a mistake earlier.Wait, if L=90, then equation B:90 = 90 / (1 + e^{-k(20 - t0)})So,1 + e^{-k(20 - t0)} = 1Thus,e^{-k(20 - t0)} = 0Which implies that -k(20 - t0) approaches negative infinity, which is only possible if k approaches infinity, which is not practical. So, this suggests that L cannot be 90.Wait, but according to our earlier steps, we arrived at L=90. So, perhaps there's an error in the process.Let me go back.We had:(L - 90)/90 = (L - 90)/18Which simplifies to:1/90 = 1/18, which is only possible if L - 90 =0, so L=90.But that leads to a contradiction because equation B would require e^{-k(20 - t0)}=0, which is impossible unless k is infinite.Therefore, perhaps my approach is flawed.Wait, maybe I made a mistake when setting up the equations.Let me try a different approach.Given the logistic function:( P(t) = frac{L}{1 + e^{-k(t - t_0)}} )We have two points: (10,50) and (20,90). Let's plug them into the equation.First, at t=10, P=50:( 50 = frac{L}{1 + e^{-k(10 - t_0)}} ) --> Equation 1At t=20, P=90:( 90 = frac{L}{1 + e^{-k(20 - t_0)}} ) --> Equation 2Let me denote ( u = e^{-k(10 - t_0)} ). Then, ( e^{-k(20 - t_0)} = e^{-k(10 - t_0)} cdot e^{-10k} = u e^{-10k} ).So, Equation 1 becomes:50 = L / (1 + u) --> 1 + u = L / 50 --> u = (L / 50) - 1Equation 2 becomes:90 = L / (1 + u e^{-10k}) --> 1 + u e^{-10k} = L / 90 --> u e^{-10k} = (L / 90) - 1Now, from Equation 1, u = (L / 50) - 1. Plug this into Equation 2:[(L / 50) - 1] e^{-10k} = (L / 90) - 1Let me denote z = e^{-10k}, so:[(L / 50) - 1] z = (L / 90) - 1So,z = [ (L / 90) - 1 ] / [ (L / 50) - 1 ]Simplify numerator and denominator:Numerator: (L - 90)/90Denominator: (L - 50)/50So,z = [ (L - 90)/90 ] / [ (L - 50)/50 ] = [ (L - 90)/90 ] * [ 50/(L - 50) ] = [50(L - 90)] / [90(L - 50)] = [5(L - 90)] / [9(L - 50)]But z = e^{-10k}, so:e^{-10k} = [5(L - 90)] / [9(L - 50)]Now, let's go back to Equation 1:u = (L / 50) - 1 = e^{-k(10 - t0)}But also, from the definition of u, u = e^{-k(10 - t0)}.Let me express t0 in terms of k.Let me take natural logarithm on both sides:ln(u) = -k(10 - t0)So,t0 = 10 - (ln(u))/kBut u = (L / 50) - 1, so:t0 = 10 - [ln( (L / 50) - 1 ) ] / kHmm, this seems complicated. Maybe I can express t0 in terms of L and k.Alternatively, let's use the fact that the logistic function is symmetric around t0. The slope is maximum at t0.But without another point, it's hard to determine t0.Wait, perhaps I can assume that the midpoint t0 is somewhere between t=10 and t=20. Let's assume t0 is between 10 and 20.But without more information, it's hard to find t0. Maybe we can express t0 in terms of L and k.Alternatively, let's consider that the logistic function passes through (10,50) and (20,90). Let's try to find L, k, t0.Let me consider that at t0, P(t0) = L/2.So, if I can find t0 such that P(t0)=L/2, but we don't have a data point at t0. However, perhaps we can use the two points to find t0.Let me denote t0 as the midpoint between t=10 and t=20 in terms of the logistic curve.But the logistic curve doesn't necessarily have t0 as the midpoint in time, but rather as the point where the growth rate is maximum.Alternatively, let's consider the ratio of the two equations.From Equation 1: 50 = L / (1 + u)From Equation 2: 90 = L / (1 + u z)Where z = e^{-10k}So, let's take the ratio of Equation 2 to Equation 1:90/50 = [1 + u] / [1 + u z]Simplify:9/5 = (1 + u) / (1 + u z)Cross-multiply:9(1 + u z) = 5(1 + u)Expand:9 + 9u z = 5 + 5uBring all terms to left:9 + 9u z - 5 - 5u = 0Simplify:4 + 9u z - 5u = 0Factor u:4 + u(9z - 5) = 0So,u(9z - 5) = -4But u = (L / 50) - 1, and z = [5(L - 90)] / [9(L - 50)]So,[(L / 50) - 1] * [9 * [5(L - 90)/9(L - 50)] - 5] = -4Simplify inside the brackets:9 * [5(L - 90)/9(L - 50)] = [5(L - 90)] / (L - 50)So,[5(L - 90)/(L - 50) - 5] = [5(L - 90) - 5(L - 50)] / (L - 50) = [5L - 450 - 5L + 250] / (L - 50) = (-200)/(L - 50)Thus,[(L / 50) - 1] * (-200)/(L - 50) = -4Simplify:[(L - 50)/50] * (-200)/(L - 50) = -4The (L - 50) terms cancel:(L - 50)/50 * (-200)/(L - 50) = (-200)/50 = -4So,-4 = -4Which is an identity, meaning our earlier steps are consistent, but it doesn't help us find L.This suggests that we need another approach because the equations are dependent and we can't solve for L uniquely without additional information.Wait, perhaps I made a wrong assumption earlier. Let me think differently.Let me consider that the logistic function is symmetric around t0. So, the time between t0 and t=10 is the same as the time between t=20 and t0 if the function is symmetric. But logistic function isn't symmetric in time; it's sigmoidal, so the slope increases to t0 and then decreases.Wait, but maybe the distances in terms of the function's values are symmetric. Let me see.At t=10, P=50; at t=20, P=90.If L is the maximum, then the distance from 50 to L is L - 50, and from 90 to L is L - 90.If the function is symmetric, the time intervals would correspond to equal fractions of the growth. But without knowing L, it's hard to say.Alternatively, let's assume that t0 is the midpoint between t=10 and t=20, so t0=15. Let's test this assumption.If t0=15, then let's see if we can find L and k.From Equation 1:50 = L / (1 + e^{-k(10 - 15)}) = L / (1 + e^{5k})From Equation 2:90 = L / (1 + e^{-k(20 - 15)}) = L / (1 + e^{-5k})Let me denote e^{5k} = m. Then, e^{-5k} = 1/m.So, Equation 1: 50 = L / (1 + m)Equation 2: 90 = L / (1 + 1/m) = L / ( (m + 1)/m ) = L m / (m + 1)From Equation 1: L = 50(1 + m)From Equation 2: 90 = [50(1 + m)] * m / (m + 1) = 50 mSo,90 = 50 mThus,m = 90 / 50 = 1.8So, m = e^{5k} = 1.8Thus,5k = ln(1.8)k = (ln(1.8))/5 ≈ (0.5878)/5 ≈ 0.1176 per minuteNow, from Equation 1:L = 50(1 + m) = 50(1 + 1.8) = 50 * 2.8 = 140So, L=140, k≈0.1176, t0=15Let me verify if this works.At t=10:P(10) = 140 / (1 + e^{-0.1176*(10 - 15)}) = 140 / (1 + e^{0.588}) ≈ 140 / (1 + 1.8) ≈ 140 / 2.8 ≈ 50 ✔️At t=20:P(20) = 140 / (1 + e^{-0.1176*(20 - 15)}) = 140 / (1 + e^{-0.588}) ≈ 140 / (1 + 0.555) ≈ 140 / 1.555 ≈ 90 ✔️Perfect! So, assuming t0=15 works, and we get consistent results.Therefore, L=140, k≈0.1176, t0=15.But let me express k more precisely.k = (ln(1.8))/5ln(1.8) ≈ 0.587787So,k ≈ 0.587787 / 5 ≈ 0.117557 per minuteWe can write it as ln(1.8)/5 or approximate it as 0.1176.Alternatively, we can express k in terms of ln(9/5)/5, since 1.8=9/5.So,k = (ln(9/5))/5Thus, the exact value is (ln(9/5))/5.So, summarizing:L=140k= (ln(9/5))/5 ≈0.1176t0=15Therefore, the values are L=140, k= (ln(9/5))/5, and t0=15.Let me check if this makes sense.At t0=15, P(t0)=140/2=70.So, at t=15, the system can handle 70 robots, which is between 50 and 90, which makes sense.The growth rate k is positive, so the function increases from 50 at t=10 to 90 at t=20, approaching L=140 asymptotically.Yes, this seems correct.So, the final answers are:1. a=1, b=1, c=12. L=140, k=(ln(9/5))/5, t0=15</think>

answer:<think>Okay, so I'm trying to solve this problem where I need to analyze the impact of community-based rehabilitation programs (CBRP) versus traditional incarceration on crime rates and community well-being. The problem is divided into two parts. Let me tackle them one by one.Part 1: Solving the Differential Equations for Crime RatesFirst, I have two differential equations for the crime rates in communities A and B. Both are linear differential equations, so I think I can solve them using integrating factors or maybe find their general solutions.Starting with Community A:The differential equation is:[frac{dC_A(t)}{dt} = -0.05C_A(t) + 2]This is a linear first-order ODE. The standard form is:[frac{dy}{dt} + P(t)y = Q(t)]So, rewriting the equation:[frac{dC_A}{dt} + 0.05C_A = 2]Here, P(t) is 0.05 and Q(t) is 2. The integrating factor (IF) is given by:[IF = e^{int P(t) dt} = e^{int 0.05 dt} = e^{0.05t}]Multiplying both sides of the ODE by the integrating factor:[e^{0.05t} frac{dC_A}{dt} + 0.05e^{0.05t}C_A = 2e^{0.05t}]The left side is the derivative of ( e^{0.05t}C_A ) with respect to t. So, integrating both sides:[int frac{d}{dt} [e^{0.05t}C_A] dt = int 2e^{0.05t} dt]Which simplifies to:[e^{0.05t}C_A = frac{2}{0.05}e^{0.05t} + C]Simplifying further:[e^{0.05t}C_A = 40e^{0.05t} + C]Divide both sides by ( e^{0.05t} ):[C_A(t) = 40 + Ce^{-0.05t}]Now, applying the initial condition ( C_A(0) = 50 ):[50 = 40 + C]So, ( C = 10 ). Therefore, the solution for Community A is:[C_A(t) = 40 + 10e^{-0.05t}]Alright, that seems straightforward. Now, moving on to Community B.The differential equation for Community B is:[frac{dC_B(t)}{dt} = -0.1C_B(t) + sin(t)]Again, this is a linear first-order ODE. Let's rewrite it in standard form:[frac{dC_B}{dt} + 0.1C_B = sin(t)]So, P(t) is 0.1 and Q(t) is sin(t). The integrating factor is:[IF = e^{int 0.1 dt} = e^{0.1t}]Multiplying both sides by the integrating factor:[e^{0.1t} frac{dC_B}{dt} + 0.1e^{0.1t}C_B = e^{0.1t}sin(t)]The left side is the derivative of ( e^{0.1t}C_B ). So, integrating both sides:[int frac{d}{dt} [e^{0.1t}C_B] dt = int e^{0.1t}sin(t) dt]This simplifies to:[e^{0.1t}C_B = int e^{0.1t}sin(t) dt + C]Now, I need to compute the integral ( int e^{0.1t}sin(t) dt ). I remember that integrals of the form ( int e^{at}sin(bt) dt ) can be solved using integration by parts twice and then solving for the integral.Let me set:Let ( u = sin(t) ), ( dv = e^{0.1t} dt )Then, ( du = cos(t) dt ), ( v = frac{1}{0.1}e^{0.1t} = 10e^{0.1t} )Integration by parts formula:[int u dv = uv - int v du]So,[int e^{0.1t}sin(t) dt = 10e^{0.1t}sin(t) - 10int e^{0.1t}cos(t) dt]Now, I need to compute ( int e^{0.1t}cos(t) dt ). Let's do integration by parts again.Let ( u = cos(t) ), ( dv = e^{0.1t} dt )Then, ( du = -sin(t) dt ), ( v = 10e^{0.1t} )So,[int e^{0.1t}cos(t) dt = 10e^{0.1t}cos(t) + 10int e^{0.1t}sin(t) dt]Now, substitute this back into the previous equation:[int e^{0.1t}sin(t) dt = 10e^{0.1t}sin(t) - 10[10e^{0.1t}cos(t) + 10int e^{0.1t}sin(t) dt]]Simplify:[int e^{0.1t}sin(t) dt = 10e^{0.1t}sin(t) - 100e^{0.1t}cos(t) - 100int e^{0.1t}sin(t) dt]Now, let's move the integral term to the left side:[int e^{0.1t}sin(t) dt + 100int e^{0.1t}sin(t) dt = 10e^{0.1t}sin(t) - 100e^{0.1t}cos(t)]Factor out the integral:[101int e^{0.1t}sin(t) dt = 10e^{0.1t}sin(t) - 100e^{0.1t}cos(t)]Divide both sides by 101:[int e^{0.1t}sin(t) dt = frac{10e^{0.1t}sin(t) - 100e^{0.1t}cos(t)}{101} + C]So, going back to the equation for ( e^{0.1t}C_B ):[e^{0.1t}C_B = frac{10e^{0.1t}sin(t) - 100e^{0.1t}cos(t)}{101} + C]Divide both sides by ( e^{0.1t} ):[C_B(t) = frac{10sin(t) - 100cos(t)}{101} + Ce^{-0.1t}]Now, apply the initial condition ( C_B(0) = 50 ):[50 = frac{10sin(0) - 100cos(0)}{101} + C]Simplify:[50 = frac{0 - 100(1)}{101} + C][50 = -frac{100}{101} + C]So, solving for C:[C = 50 + frac{100}{101} = frac{5050}{101} + frac{100}{101} = frac{5150}{101} approx 51.0]Wait, let me compute that exactly:50 is 5050/101, so 5050 + 100 = 5150, so C = 5150/101.So, the solution for Community B is:[C_B(t) = frac{10sin(t) - 100cos(t)}{101} + frac{5150}{101}e^{-0.1t}]Simplify the constants:[C_B(t) = frac{10}{101}sin(t) - frac{100}{101}cos(t) + frac{5150}{101}e^{-0.1t}]I can factor out 10/101 from the first two terms:[C_B(t) = frac{10}{101}(sin(t) - 10cos(t)) + frac{5150}{101}e^{-0.1t}]Alternatively, I can write 5150/101 as 51.0 approximately, but since 101*51 = 5151, so 5150 is 51 - 1/101, so it's approximately 50.9901.But maybe it's better to keep it as a fraction for exactness.So, summarizing:- Community A: ( C_A(t) = 40 + 10e^{-0.05t} )- Community B: ( C_B(t) = frac{10}{101}sin(t) - frac{100}{101}cos(t) + frac{5150}{101}e^{-0.1t} )Part 2: Evaluating the Community Well-Being Index (WBI) for Community BThe WBI is given by:[W_B(t) = int_0^t e^{-0.01u} (50 - C_B(u)) , du]We need to substitute ( C_B(u) ) into this integral and evaluate it at t = 12.First, let's write out ( 50 - C_B(u) ):[50 - C_B(u) = 50 - left( frac{10}{101}sin(u) - frac{100}{101}cos(u) + frac{5150}{101}e^{-0.1u} right )]Simplify:[50 - C_B(u) = 50 - frac{10}{101}sin(u) + frac{100}{101}cos(u) - frac{5150}{101}e^{-0.1u}]So, plugging this into the integral:[W_B(t) = int_0^t e^{-0.01u} left( 50 - frac{10}{101}sin(u) + frac{100}{101}cos(u) - frac{5150}{101}e^{-0.1u} right ) du]This integral can be split into four separate integrals:[W_B(t) = 50 int_0^t e^{-0.01u} du - frac{10}{101} int_0^t e^{-0.01u}sin(u) du + frac{100}{101} int_0^t e^{-0.01u}cos(u) du - frac{5150}{101} int_0^t e^{-0.01u}e^{-0.1u} du]Simplify each integral one by one.1. First integral:[I_1 = 50 int_0^t e^{-0.01u} du]Integrate:[I_1 = 50 left[ frac{e^{-0.01u}}{-0.01} right ]_0^t = 50 left( frac{1 - e^{-0.01t}}{0.01} right ) = 50 times 100 (1 - e^{-0.01t}) = 5000(1 - e^{-0.01t})]2. Second integral:[I_2 = - frac{10}{101} int_0^t e^{-0.01u}sin(u) du]This integral is similar to the one we did earlier. Let me recall the formula for ( int e^{at}sin(bt) dt ). The general formula is:[int e^{at}sin(bt) dt = frac{e^{at}}{a^2 + b^2}(asin(bt) - bcos(bt)) + C]In our case, a = -0.01 and b = 1. So,[int e^{-0.01u}sin(u) du = frac{e^{-0.01u}}{(-0.01)^2 + 1^2} (-0.01sin(u) - cos(u)) + C]Simplify denominator:[(-0.01)^2 + 1 = 0.0001 + 1 = 1.0001]So,[int e^{-0.01u}sin(u) du = frac{e^{-0.01u}}{1.0001} (-0.01sin(u) - cos(u)) + C]Therefore, evaluating from 0 to t:[I_2 = - frac{10}{101} left[ frac{e^{-0.01t}}{1.0001} (-0.01sin(t) - cos(t)) - frac{1}{1.0001} ( -0.01sin(0) - cos(0) ) right ]]Simplify:[I_2 = - frac{10}{101} left[ frac{e^{-0.01t}(-0.01sin(t) - cos(t))}{1.0001} - frac{0 - 1}{1.0001} right ]][I_2 = - frac{10}{101} left[ frac{ -0.01sin(t) e^{-0.01t} - cos(t) e^{-0.01t} + 1 }{1.0001} right ]]Factor out the negative sign:[I_2 = - frac{10}{101} times frac{ -0.01sin(t) e^{-0.01t} - cos(t) e^{-0.01t} + 1 }{1.0001 }][I_2 = frac{10}{101 times 1.0001} [0.01sin(t) e^{-0.01t} + cos(t) e^{-0.01t} - 1]]Note that 101 * 1.0001 = 101.0101, but maybe we can keep it as fractions for exactness.Wait, 1.0001 is 10001/10000. So, 101 * 1.0001 = 101 * 10001/10000 = (101*10001)/10000.But perhaps it's better to just keep it as 101*1.0001 for now.3. Third integral:[I_3 = frac{100}{101} int_0^t e^{-0.01u}cos(u) du]Again, using the standard integral formula:[int e^{at}cos(bt) dt = frac{e^{at}}{a^2 + b^2}(acos(bt) + bsin(bt)) + C]Here, a = -0.01, b = 1. So,[int e^{-0.01u}cos(u) du = frac{e^{-0.01u}}{(-0.01)^2 + 1^2} (-0.01cos(u) + sin(u)) + C]Which is:[frac{e^{-0.01u}}{1.0001} (-0.01cos(u) + sin(u)) + C]Evaluating from 0 to t:[I_3 = frac{100}{101} left[ frac{e^{-0.01t}(-0.01cos(t) + sin(t))}{1.0001} - frac{(-0.01cos(0) + sin(0))}{1.0001} right ]]Simplify:[I_3 = frac{100}{101} left[ frac{ -0.01cos(t) e^{-0.01t} + sin(t) e^{-0.01t} }{1.0001} - frac{ -0.01(1) + 0 }{1.0001} right ]][I_3 = frac{100}{101} left[ frac{ -0.01cos(t) e^{-0.01t} + sin(t) e^{-0.01t} + 0.01 }{1.0001} right ]]Factor out e^{-0.01t}:[I_3 = frac{100}{101 times 1.0001} [ -0.01cos(t) + sin(t) ] e^{-0.01t} + frac{100}{101} times frac{0.01}{1.0001}]Wait, actually, let me re-express it properly:[I_3 = frac{100}{101} times frac{1}{1.0001} [ (-0.01cos(t) + sin(t)) e^{-0.01t} + 0.01 ]]So, combining terms:[I_3 = frac{100}{101 times 1.0001} [ (-0.01cos(t) + sin(t)) e^{-0.01t} + 0.01 ]]4. Fourth integral:[I_4 = - frac{5150}{101} int_0^t e^{-0.01u} e^{-0.1u} du = - frac{5150}{101} int_0^t e^{-0.11u} du]Simplify the exponent:[-0.01 - 0.1 = -0.11]So,[I_4 = - frac{5150}{101} left[ frac{e^{-0.11u}}{-0.11} right ]_0^t = - frac{5150}{101} left( frac{1 - e^{-0.11t}}{0.11} right )]Simplify:[I_4 = - frac{5150}{101 times 0.11} (1 - e^{-0.11t})]Calculate 5150 / (101 * 0.11):First, 101 * 0.11 = 11.11So, 5150 / 11.11 ≈ 463.636But let's compute it exactly:5150 / 11.11 = 515000 / 1111 ≈ 463.636But maybe we can keep it as a fraction:5150 / 11.11 = 515000 / 1111But 515000 divided by 1111: Let's see, 1111 * 463 = 1111*400=444400, 1111*60=66660, 1111*3=3333. So total 444400 + 66660 = 511060 + 3333 = 514,393. So 1111*463 = 514,393. Then 515,000 - 514,393 = 607. So 515000 / 1111 = 463 + 607/1111 ≈ 463.546But perhaps it's better to leave it as 5150 / 11.11 for now.So, putting it all together, the WBI is:[W_B(t) = I_1 + I_2 + I_3 + I_4]Which is:[W_B(t) = 5000(1 - e^{-0.01t}) + frac{10}{101 times 1.0001} [0.01sin(t) e^{-0.01t} + cos(t) e^{-0.01t} - 1] + frac{100}{101 times 1.0001} [ (-0.01cos(t) + sin(t)) e^{-0.01t} + 0.01 ] - frac{5150}{101 times 0.11} (1 - e^{-0.11t})]This expression is quite complex. Maybe we can combine like terms.First, let me note that 101 * 1.0001 = 101.0101, and 101 * 0.11 = 11.11.Let me compute the coefficients:For I_2:Coefficient: 10 / (101 * 1.0001) ≈ 10 / 101.0101 ≈ 0.0990099For I_3:Coefficient: 100 / (101 * 1.0001) ≈ 100 / 101.0101 ≈ 0.990099For I_4:Coefficient: -5150 / (101 * 0.11) ≈ -5150 / 11.11 ≈ -463.636Now, let's see if we can combine the terms involving e^{-0.01t}:From I_2:0.01 sin(t) e^{-0.01t} + cos(t) e^{-0.01t}From I_3:(-0.01 cos(t) + sin(t)) e^{-0.01t}So, combining these:[0.01 sin(t) + cos(t) - 0.01 cos(t) + sin(t)] e^{-0.01t}Simplify:(0.01 sin(t) + sin(t)) + (cos(t) - 0.01 cos(t)) = (1.01 sin(t)) + (0.99 cos(t))So, the combined term is:(1.01 sin(t) + 0.99 cos(t)) e^{-0.01t}Now, the coefficients from I_2 and I_3 are 0.0990099 and 0.990099 respectively. So, multiplying:From I_2: 0.0990099 * (1.01 sin(t) + 0.99 cos(t)) e^{-0.01t}From I_3: 0.990099 * (1.01 sin(t) + 0.99 cos(t)) e^{-0.01t}Wait, actually, no. Wait, I think I made a mistake in combining the terms. Let me re-examine.Wait, in I_2, the coefficient is 10/(101*1.0001) ≈ 0.0990099, and the expression inside is [0.01 sin(t) e^{-0.01t} + cos(t) e^{-0.01t} - 1]. Similarly, in I_3, the coefficient is 100/(101*1.0001) ≈ 0.990099, and the expression is [ (-0.01 cos(t) + sin(t)) e^{-0.01t} + 0.01 ].So, when combining the e^{-0.01t} terms:From I_2: 0.0990099*(0.01 sin(t) + cos(t)) e^{-0.01t}From I_3: 0.990099*(-0.01 cos(t) + sin(t)) e^{-0.01t}So, combining these:0.0990099*(0.01 sin(t) + cos(t)) + 0.990099*(-0.01 cos(t) + sin(t)) all multiplied by e^{-0.01t}Let me compute each part:First term: 0.0990099*0.01 sin(t) = 0.000990099 sin(t)Second term: 0.0990099*cos(t)Third term: 0.990099*(-0.01) cos(t) = -0.00990099 cos(t)Fourth term: 0.990099*sin(t)So, combining like terms:sin(t): 0.000990099 + 0.990099 ≈ 0.991089099 sin(t)cos(t): 0.0990099 - 0.00990099 ≈ 0.08910891 cos(t)So, the combined coefficient for e^{-0.01t} is approximately (0.991089099 sin(t) + 0.08910891 cos(t)).Now, the constants from I_2 and I_3:From I_2: 0.0990099*(-1) = -0.0990099From I_3: 0.990099*0.01 = 0.00990099So, combining constants:-0.0990099 + 0.00990099 ≈ -0.08910891So, putting it all together, W_B(t) is approximately:5000(1 - e^{-0.01t}) + (0.991089099 sin(t) + 0.08910891 cos(t)) e^{-0.01t} - 0.08910891 - 463.636(1 - e^{-0.11t})Simplify further:Combine the constants: -0.08910891 - 463.636 ≈ -463.725So,[W_B(t) ≈ 5000(1 - e^{-0.01t}) + (0.991089099 sin(t) + 0.08910891 cos(t)) e^{-0.01t} - 463.725(1 - e^{-0.11t})]But this is an approximate expression. Alternatively, maybe we can keep it in terms of exact fractions.Wait, let me try to express it more precisely.Let me note that 101 * 1.0001 = 101 * (10001/10000) = (101*10001)/10000.Similarly, 101 * 0.11 = 11.11.But perhaps it's better to compute each term numerically for t = 12.Given that t = 12, let's compute each integral numerically.First, compute I_1 at t=12:I_1 = 5000(1 - e^{-0.01*12}) = 5000(1 - e^{-0.12})Compute e^{-0.12} ≈ 0.88692044So, I_1 ≈ 5000*(1 - 0.88692044) ≈ 5000*0.11307956 ≈ 565.3978Next, compute I_2 at t=12:I_2 = - (10/101) * [ integral from 0 to 12 of e^{-0.01u} sin(u) du ]But we already have the expression for the integral:[int e^{-0.01u}sin(u) du = frac{e^{-0.01u}}{1.0001}(-0.01sin(u) - cos(u)) + C]Evaluated from 0 to 12:[left[ frac{e^{-0.12}}{1.0001}(-0.01sin(12) - cos(12)) - frac{1}{1.0001}(0 - 1) right ]]Compute each part:First, compute e^{-0.12} ≈ 0.88692044Compute sin(12) and cos(12). Note that 12 radians is approximately 687.549 degrees, which is more than 2π (≈6.283). So, 12 radians is 12 - 2π*1 ≈ 12 - 6.283 ≈ 5.717 radians, which is still more than π. 5.717 - π ≈ 5.717 - 3.142 ≈ 2.575 radians. So, it's in the third quadrant.But regardless, let's compute sin(12) and cos(12) using calculator:sin(12) ≈ -0.536572918cos(12) ≈ -0.843853959So, plug these in:First term:0.88692044 / 1.0001 * (-0.01*(-0.536572918) - (-0.843853959))= 0.88692044 / 1.0001 * (0.005365729 + 0.843853959)= 0.88692044 / 1.0001 * 0.849219688≈ 0.88692044 * 0.849219688 / 1.0001≈ 0.75275 / 1.0001 ≈ 0.75268Second term:- [ ( -0.01*0 - 1 ) / 1.0001 ] = - [ (0 - 1)/1.0001 ] = - [ -1 / 1.0001 ] = 1 / 1.0001 ≈ 0.99990001So, the entire expression inside the brackets is:0.75268 + 0.99990001 ≈ 1.75258Therefore, I_2 ≈ - (10/101) * 1.75258 ≈ -0.10 * 1.75258 ≈ -0.175258Wait, 10/101 ≈ 0.0990099, so 0.0990099 * 1.75258 ≈ 0.1737So, I_2 ≈ -0.1737Next, compute I_3 at t=12:I_3 = (100/101) * [ integral from 0 to 12 of e^{-0.01u} cos(u) du ]Using the integral formula:[int e^{-0.01u}cos(u) du = frac{e^{-0.01u}}{1.0001}(-0.01cos(u) + sin(u)) + C]Evaluated from 0 to 12:[left[ frac{e^{-0.12}}{1.0001}(-0.01cos(12) + sin(12)) - frac{1}{1.0001}(-0.01cos(0) + sin(0)) right ]]Compute each part:First term:0.88692044 / 1.0001 * (-0.01*(-0.843853959) + (-0.536572918))= 0.88692044 / 1.0001 * (0.0084385396 - 0.536572918)= 0.88692044 / 1.0001 * (-0.528134378)≈ 0.88692044 * (-0.528134378) / 1.0001≈ -0.4684 / 1.0001 ≈ -0.4683Second term:- [ (-0.01*1 + 0 ) / 1.0001 ] = - [ (-0.01)/1.0001 ] = 0.01 / 1.0001 ≈ 0.009999So, the entire expression inside the brackets is:-0.4683 + 0.009999 ≈ -0.4583Therefore, I_3 ≈ (100/101) * (-0.4583) ≈ 0.990099 * (-0.4583) ≈ -0.4544Finally, compute I_4 at t=12:I_4 = - (5150 / 101) * (1 - e^{-0.11*12}) / 0.11= - (5150 / 101) * (1 - e^{-1.32}) / 0.11Compute e^{-1.32} ≈ 0.266641So, 1 - 0.266641 ≈ 0.733359Compute 5150 / 101 ≈ 51.0So, I_4 ≈ -51.0 * 0.733359 / 0.11 ≈ -51.0 * 6.6669 ≈ -51.0 * 6.6669 ≈ -340.0Wait, let's compute it step by step:5150 / 101 = 51.0 (exactly, since 101*51=5151, so 5150/101=51 - 1/101≈50.9901)So, 5150 / 101 ≈ 50.9901Then, (1 - e^{-1.32}) ≈ 0.733359So, 50.9901 * 0.733359 ≈ 50.9901 * 0.733359 ≈ 37.41Then, divide by 0.11: 37.41 / 0.11 ≈ 340.09So, I_4 ≈ -340.09Now, summing up all the integrals:I_1 ≈ 565.3978I_2 ≈ -0.1737I_3 ≈ -0.4544I_4 ≈ -340.09So, total W_B(12) ≈ 565.3978 - 0.1737 - 0.4544 - 340.09 ≈Compute step by step:565.3978 - 0.1737 ≈ 565.2241565.2241 - 0.4544 ≈ 564.7697564.7697 - 340.09 ≈ 224.6797So, approximately 224.68But let me check the calculations again because I might have made an error in signs or coefficients.Wait, in I_4, the formula is:I_4 = - (5150 / 101) * (1 - e^{-0.11t}) / 0.11So, plugging t=12:= - (5150 / 101) * (1 - e^{-1.32}) / 0.11Compute 5150 / 101 ≈ 51.01 - e^{-1.32} ≈ 0.733359So, 51.0 * 0.733359 ≈ 37.4013Divide by 0.11: 37.4013 / 0.11 ≈ 340.012So, I_4 ≈ -340.012So, total W_B(12) ≈ 565.3978 - 0.1737 - 0.4544 - 340.012 ≈565.3978 - 0.1737 = 565.2241565.2241 - 0.4544 = 564.7697564.7697 - 340.012 ≈ 224.7577So, approximately 224.76But let me verify the I_2 and I_3 calculations because they are small compared to I_1 and I_4.I_2 ≈ -0.1737I_3 ≈ -0.4544So, total of I_2 + I_3 ≈ -0.6281So, total W_B(12) ≈ 565.3978 - 0.6281 - 340.012 ≈ 565.3978 - 340.6401 ≈ 224.7577Yes, that seems consistent.So, the WBI at t=12 is approximately 224.76.But let me check if I made any mistake in the signs.Looking back:I_2 was computed as - (10/101) * [ integral result ]The integral result was approximately 1.75258, so I_2 ≈ -0.1737I_3 was computed as (100/101) * [ integral result ]The integral result was approximately -0.4583, so I_3 ≈ -0.4544I_4 was computed as -340.012I_1 was 565.3978So, adding them up: 565.3978 - 0.1737 - 0.4544 - 340.012 ≈ 224.76Therefore, W_B(12) ≈ 224.76But let me see if I can compute it more accurately.Alternatively, maybe I should compute each term more precisely.But given the time constraints, I think 224.76 is a reasonable approximation.Final AnswerThe crime rates are ( C_A(t) = 40 + 10e^{-0.05t} ) and ( C_B(t) = frac{10}{101}sin(t) - frac{100}{101}cos(t) + frac{5150}{101}e^{-0.1t} ). The community well-being index at 12 months is approximately boxed{224.76}.</think>

answer:<think>Okay, so I need to help Dr. Smith with her novel by solving this partial differential equation related to the diffusion of a medication. Let me start by understanding the problem step by step.The equation given is:[ frac{partial C}{partial t} = D frac{partial^2 C}{partial x^2} - kC ]This looks like a reaction-diffusion equation. The term ( D frac{partial^2 C}{partial x^2} ) represents the diffusion of the medication, and the term ( -kC ) represents the metabolism or degradation of the medication over time.The initial condition is:[ C(x,0) = C_0 e^{-alpha x^2} ]So, at time t=0, the concentration is a Gaussian distribution centered at x=0 with width determined by α.I need to find the general solution for C(x,t). Hmm, partial differential equations can be tricky, but I remember that for linear PDEs, we can sometimes use methods like separation of variables or Fourier transforms. Since the equation is linear and has constant coefficients, maybe Fourier transforms are the way to go.Let me recall that the Fourier transform can convert a PDE into an ordinary differential equation (ODE), which is easier to solve. So, let's take the Fourier transform of both sides.First, define the Fourier transform of C(x,t) as:[ mathcal{F}{C(x,t)} = hat{C}(k,t) = int_{-infty}^{infty} C(x,t) e^{-ikx} dx ]Taking the Fourier transform of the PDE:[ mathcal{F}left{frac{partial C}{partial t}right} = mathcal{F}left{D frac{partial^2 C}{partial x^2} - kCright} ]The left side becomes:[ frac{partial hat{C}}{partial t} ]The right side, using the properties of Fourier transforms:- The Fourier transform of ( frac{partial^2 C}{partial x^2} ) is ( -k^2 hat{C} )- The Fourier transform of ( -kC ) is ( -k hat{C} )So, putting it together:[ frac{partial hat{C}}{partial t} = -D k^2 hat{C} - k hat{C} ]This simplifies to:[ frac{partial hat{C}}{partial t} = - (D k^2 + k) hat{C} ]This is an ODE in terms of t. The solution to this ODE is:[ hat{C}(k,t) = hat{C}(k,0) e^{ - (D k^2 + k) t } ]Now, we need the initial condition in Fourier space. The initial condition is:[ C(x,0) = C_0 e^{-alpha x^2} ]Taking the Fourier transform of this:[ hat{C}(k,0) = int_{-infty}^{infty} C_0 e^{-alpha x^2} e^{-ikx} dx ]I remember that the Fourier transform of a Gaussian is another Gaussian. The formula is:[ mathcal{F}{e^{-alpha x^2}} = sqrt{frac{pi}{alpha}} e^{-k^2 / (4alpha)} ]So, applying this:[ hat{C}(k,0) = C_0 sqrt{frac{pi}{alpha}} e^{-k^2 / (4alpha)} ]Therefore, the solution in Fourier space is:[ hat{C}(k,t) = C_0 sqrt{frac{pi}{alpha}} e^{-k^2 / (4alpha)} e^{ - (D k^2 + k) t } ]Now, to find C(x,t), we need to take the inverse Fourier transform:[ C(x,t) = frac{1}{2pi} int_{-infty}^{infty} hat{C}(k,t) e^{ikx} dk ]Substituting the expression for ( hat{C}(k,t) ):[ C(x,t) = frac{C_0 sqrt{frac{pi}{alpha}}}{2pi} int_{-infty}^{infty} e^{-k^2 / (4alpha)} e^{ - (D k^2 + k) t } e^{ikx} dk ]Simplify the constants:[ C(x,t) = frac{C_0}{2 sqrt{alpha pi}} int_{-infty}^{infty} e^{-k^2 / (4alpha)} e^{ -D k^2 t - k t } e^{ikx} dk ]Combine the exponents:The exponent terms are:- ( -k^2 / (4alpha) )- ( -D k^2 t )- ( -k t )- ( ikx )Combine the quadratic terms:[ -k^2 left( frac{1}{4alpha} + D t right) ]And the linear term:[ -k t + ikx = k (ix - t) ]So, the exponent becomes:[ -k^2 left( frac{1}{4alpha} + D t right) + k (ix - t) ]This is a quadratic in k, so the integral is a Gaussian integral. The integral of a Gaussian of the form:[ int_{-infty}^{infty} e^{-a k^2 + b k} dk = sqrt{frac{pi}{a}} e^{b^2 / (4a)} ]Where ( a = frac{1}{4alpha} + D t ) and ( b = ix - t ).So, applying this formula:[ int_{-infty}^{infty} e^{-k^2 left( frac{1}{4alpha} + D t right) + k (ix - t)} dk = sqrt{frac{pi}{frac{1}{4alpha} + D t}} e^{(ix - t)^2 / (4 (frac{1}{4alpha} + D t))} ]Simplify the denominator:[ frac{1}{4alpha} + D t = frac{1 + 4 alpha D t}{4 alpha} ]So, the square root term becomes:[ sqrt{frac{pi}{frac{1 + 4 alpha D t}{4 alpha}}} = sqrt{frac{4 alpha pi}{1 + 4 alpha D t}} ]And the exponent term:[ frac{(ix - t)^2}{4 (frac{1 + 4 alpha D t}{4 alpha})} = frac{(ix - t)^2 alpha}{1 + 4 alpha D t} ]Putting it all together, the integral becomes:[ sqrt{frac{4 alpha pi}{1 + 4 alpha D t}} e^{ frac{alpha (ix - t)^2}{1 + 4 alpha D t} } ]Therefore, substituting back into C(x,t):[ C(x,t) = frac{C_0}{2 sqrt{alpha pi}} times sqrt{frac{4 alpha pi}{1 + 4 alpha D t}} e^{ frac{alpha (ix - t)^2}{1 + 4 alpha D t} } ]Simplify the constants:The constants multiply as:[ frac{C_0}{2 sqrt{alpha pi}} times sqrt{frac{4 alpha pi}{1 + 4 alpha D t}} = frac{C_0}{2 sqrt{alpha pi}} times frac{2 sqrt{alpha pi}}{sqrt{1 + 4 alpha D t}} = frac{C_0}{sqrt{1 + 4 alpha D t}} ]So, the concentration becomes:[ C(x,t) = frac{C_0}{sqrt{1 + 4 alpha D t}} e^{ frac{alpha (ix - t)^2}{1 + 4 alpha D t} } ]Wait, hold on. The exponent has an imaginary term because of the ix. But concentration should be a real function. Hmm, maybe I made a mistake in handling the exponent.Let me check. The exponent was:[ frac{alpha (ix - t)^2}{1 + 4 alpha D t} ]Let me expand (ix - t)^2:[ (ix - t)^2 = (ix)^2 - 2 ix t + t^2 = -x^2 - 2 i x t + t^2 ]So, substituting back:[ frac{alpha (-x^2 - 2 i x t + t^2)}{1 + 4 alpha D t} ]So, the exponent is:[ frac{alpha (-x^2 + t^2)}{1 + 4 alpha D t} - frac{2 i alpha x t}{1 + 4 alpha D t} ]Therefore, the exponential term can be written as:[ e^{ frac{alpha (-x^2 + t^2)}{1 + 4 alpha D t} } e^{ - frac{2 i alpha x t}{1 + 4 alpha D t} } ]But since C(x,t) is real, the imaginary part must cancel out. Wait, but in our solution, we have an imaginary exponential. That suggests that perhaps I made a mistake in the Fourier transform approach.Alternatively, maybe I should have considered that the solution is real, so perhaps I need to take the real part or something. Hmm, maybe I should have used a symmetric Fourier transform or considered that the solution is real.Wait, another approach: perhaps I should have used the method of eigenfunction expansion or separation of variables. Let me think.Alternatively, maybe I can write the solution as a Gaussian function that evolves over time, considering both diffusion and decay.Wait, let's consider that the equation is linear, so the solution can be expressed as a convolution of the initial condition with the Green's function of the operator.The Green's function G(x,t) satisfies:[ frac{partial G}{partial t} = D frac{partial^2 G}{partial x^2} - k G ]With the initial condition G(x,0) = δ(x).So, if I can find G(x,t), then the solution C(x,t) is the convolution of G(x,t) with the initial condition C(x,0).So, let's try to find G(x,t). Taking Fourier transform in x:[ frac{partial hat{G}}{partial t} = -D k^2 hat{G} - k hat{G} ]This is the same ODE as before. So,[ hat{G}(k,t) = e^{ - (D k^2 + k) t } ]Because the initial condition is δ(x), whose Fourier transform is 1.Therefore, G(x,t) is the inverse Fourier transform of ( e^{ - (D k^2 + k) t } ).So, similar to before, G(x,t) would be:[ G(x,t) = frac{1}{2pi} int_{-infty}^{infty} e^{ - (D k^2 + k) t } e^{ikx} dk ]Which is the same integral as before, except without the initial condition factor. So, perhaps I can use the same approach.So, combining the exponent:[ -D k^2 t - k t + ikx ]Which is:[ -k^2 (D t) - k (t - i x) ]This is a quadratic in k, so again, completing the square or using the Gaussian integral formula.Let me write it as:[ -D t left( k^2 + frac{k (t - i x)}{D t} right) ]Wait, maybe better to factor out D t:[ -D t left( k^2 + frac{k (t - i x)}{D t} right) = -D t left( k^2 + frac{k (t - i x)}{D t} right) ]Let me complete the square inside the parentheses:Let me denote A = 1, B = (t - i x)/(D t)So, the expression is:[ k^2 + B k = left( k + frac{B}{2} right)^2 - frac{B^2}{4} ]Therefore,[ -D t left( left( k + frac{B}{2} right)^2 - frac{B^2}{4} right ) = -D t left( k + frac{B}{2} right)^2 + frac{D t B^2}{4} ]So, the integral becomes:[ frac{1}{2pi} e^{ frac{D t B^2}{4} } int_{-infty}^{infty} e^{ -D t left( k + frac{B}{2} right)^2 } dk ]The integral is a Gaussian integral:[ int_{-infty}^{infty} e^{-a (k + c)^2} dk = sqrt{frac{pi}{a}} ]So, here a = D t, so the integral is:[ sqrt{frac{pi}{D t}} ]Therefore, putting it all together:[ G(x,t) = frac{1}{2pi} e^{ frac{D t B^2}{4} } sqrt{frac{pi}{D t}} ]Simplify:[ G(x,t) = frac{1}{2 sqrt{pi D t}} e^{ frac{D t B^2}{4} } ]Now, substitute back B:[ B = frac{t - i x}{D t} ]So,[ B^2 = left( frac{t - i x}{D t} right)^2 = frac{t^2 - 2 i t x - x^2}{D^2 t^2} ]Therefore,[ frac{D t B^2}{4} = frac{D t (t^2 - 2 i t x - x^2)}{4 D^2 t^2} = frac{t^2 - 2 i t x - x^2}{4 D t} ]So, the exponent becomes:[ frac{t^2 - 2 i t x - x^2}{4 D t} ]Therefore, G(x,t) is:[ G(x,t) = frac{1}{2 sqrt{pi D t}} e^{ frac{t^2 - 2 i t x - x^2}{4 D t} } ]Simplify the exponent:[ frac{t^2 - x^2}{4 D t} - frac{2 i t x}{4 D t} = frac{t^2 - x^2}{4 D t} - frac{i x}{2 D} ]So,[ G(x,t) = frac{1}{2 sqrt{pi D t}} e^{ frac{t^2 - x^2}{4 D t} } e^{ - frac{i x}{2 D} } ]But this still has an imaginary component. However, since the Green's function should be real, perhaps I made a mistake in the calculation. Alternatively, maybe the Green's function isn't real because of the complex exponent, but when convolved with the initial condition, the result is real.Wait, perhaps I should have considered that the Fourier transform approach inherently includes complex exponentials, but the final solution is real. So, maybe I need to take the real part or something. Alternatively, perhaps I should have used a different method.Wait, another thought: maybe I can write the solution as a product of a Gaussian and an exponential decay term. Let me think about the form.Given the PDE:[ frac{partial C}{partial t} = D frac{partial^2 C}{partial x^2} - k C ]This is similar to a heat equation with a decay term. So, perhaps the solution is a Gaussian that spreads out due to diffusion and decays exponentially due to metabolism.So, maybe the solution is:[ C(x,t) = frac{C_0}{sqrt{1 + 4 alpha D t}} e^{ - frac{alpha x^2}{1 + 4 alpha D t} } e^{ -k t } ]Wait, but that seems too simplistic. Let me check the dimensions. The exponent in the Gaussian should have units of inverse length squared. Let's see:In the initial condition, we have ( e^{-alpha x^2} ), so α has units of inverse length squared.In the solution, the exponent is ( - frac{alpha x^2}{1 + 4 alpha D t} ). The denominator has units of inverse length squared times time times time (since D has units of length squared per time). Wait, D has units of m²/s, so 4 α D t has units of (1/m²)(m²/s)(s) = 1/m² * m² = 1. So, the denominator is dimensionless? Wait, no:Wait, α is 1/m², D is m²/s, t is s. So, 4 α D t is (1/m²)(m²/s)(s) = 1. So, yes, the denominator is dimensionless, so the exponent is dimensionless.Similarly, the term ( e^{-k t} ) is dimensionless because k has units of 1/s, t is s.So, the overall solution would have units of concentration, which is correct.But does this form satisfy the PDE?Let me test it. Let me denote:[ C(x,t) = frac{C_0}{sqrt{1 + 4 alpha D t}} e^{ - frac{alpha x^2}{1 + 4 alpha D t} } e^{ -k t } ]Let me compute the partial derivatives.First, compute ( frac{partial C}{partial t} ):Let me denote ( A = sqrt{1 + 4 alpha D t} ), so ( A = (1 + 4 alpha D t)^{1/2} )Then, ( frac{1}{A} = (1 + 4 alpha D t)^{-1/2} )So, ( frac{partial}{partial t} left( frac{1}{A} right ) = - frac{1}{2} (1 + 4 alpha D t)^{-3/2} cdot 4 alpha D = - frac{2 alpha D}{(1 + 4 alpha D t)^{3/2}} )Next, the exponential term ( e^{ - frac{alpha x^2}{1 + 4 alpha D t} } ). Let me denote ( B = - frac{alpha x^2}{1 + 4 alpha D t} ), so ( frac{partial B}{partial t} = - alpha x^2 cdot frac{-4 alpha D}{(1 + 4 alpha D t)^2} = frac{4 alpha^2 D x^2}{(1 + 4 alpha D t)^2} )Therefore, the derivative of the exponential term is ( e^{B} cdot frac{partial B}{partial t} = e^{ - frac{alpha x^2}{1 + 4 alpha D t} } cdot frac{4 alpha^2 D x^2}{(1 + 4 alpha D t)^2} )Also, the term ( e^{-k t} ) has derivative ( -k e^{-k t} )Putting it all together:[ frac{partial C}{partial t} = left( - frac{2 alpha D}{(1 + 4 alpha D t)^{3/2}} right ) e^{ - frac{alpha x^2}{1 + 4 alpha D t} } e^{-k t} + frac{1}{sqrt{1 + 4 alpha D t}} e^{ - frac{alpha x^2}{1 + 4 alpha D t} } cdot frac{4 alpha^2 D x^2}{(1 + 4 alpha D t)^2} e^{-k t} + frac{1}{sqrt{1 + 4 alpha D t}} e^{ - frac{alpha x^2}{1 + 4 alpha D t} } cdot (-k) e^{-k t} ]Simplify each term:First term:[ - frac{2 alpha D}{(1 + 4 alpha D t)^{3/2}} e^{ - frac{alpha x^2}{1 + 4 alpha D t} } e^{-k t} ]Second term:[ frac{4 alpha^2 D x^2}{(1 + 4 alpha D t)^{5/2}} e^{ - frac{alpha x^2}{1 + 4 alpha D t} } e^{-k t} ]Third term:[ - frac{k}{sqrt{1 + 4 alpha D t}} e^{ - frac{alpha x^2}{1 + 4 alpha D t} } e^{-k t} ]Now, let's compute ( D frac{partial^2 C}{partial x^2} - k C )First, compute ( frac{partial C}{partial x} ):Let me denote ( C = frac{C_0}{A} e^{B} e^{-k t} ), where ( A = sqrt{1 + 4 alpha D t} ), ( B = - frac{alpha x^2}{1 + 4 alpha D t} )So,[ frac{partial C}{partial x} = frac{C_0}{A} e^{B} e^{-k t} cdot frac{partial B}{partial x} = frac{C_0}{A} e^{B} e^{-k t} cdot left( - frac{2 alpha x}{1 + 4 alpha D t} right ) ]Then, ( frac{partial^2 C}{partial x^2} ):[ frac{partial}{partial x} left( - frac{2 alpha x}{1 + 4 alpha D t} cdot frac{C_0}{A} e^{B} e^{-k t} right ) ]This is:[ - frac{2 alpha}{1 + 4 alpha D t} cdot frac{C_0}{A} e^{B} e^{-k t} - frac{2 alpha x}{1 + 4 alpha D t} cdot frac{C_0}{A} e^{B} e^{-k t} cdot frac{partial B}{partial x} ]Simplify:First term:[ - frac{2 alpha}{1 + 4 alpha D t} cdot frac{C_0}{A} e^{B} e^{-k t} ]Second term:[ - frac{2 alpha x}{1 + 4 alpha D t} cdot frac{C_0}{A} e^{B} e^{-k t} cdot left( - frac{2 alpha x}{1 + 4 alpha D t} right ) = frac{4 alpha^2 x^2}{(1 + 4 alpha D t)^2} cdot frac{C_0}{A} e^{B} e^{-k t} ]Therefore, ( frac{partial^2 C}{partial x^2} = left( - frac{2 alpha}{1 + 4 alpha D t} + frac{4 alpha^2 x^2}{(1 + 4 alpha D t)^2} right ) cdot frac{C_0}{A} e^{B} e^{-k t} )Multiply by D:[ D frac{partial^2 C}{partial x^2} = D left( - frac{2 alpha}{1 + 4 alpha D t} + frac{4 alpha^2 x^2}{(1 + 4 alpha D t)^2} right ) cdot frac{C_0}{A} e^{B} e^{-k t} ]Now, subtract kC:[ D frac{partial^2 C}{partial x^2} - k C = D left( - frac{2 alpha}{1 + 4 alpha D t} + frac{4 alpha^2 x^2}{(1 + 4 alpha D t)^2} right ) cdot frac{C_0}{A} e^{B} e^{-k t} - k cdot frac{C_0}{A} e^{B} e^{-k t} ]Factor out ( frac{C_0}{A} e^{B} e^{-k t} ):[ left( D left( - frac{2 alpha}{1 + 4 alpha D t} + frac{4 alpha^2 x^2}{(1 + 4 alpha D t)^2} right ) - k right ) cdot frac{C_0}{A} e^{B} e^{-k t} ]Now, let's compare this with ( frac{partial C}{partial t} ):Earlier, we had:[ frac{partial C}{partial t} = left( - frac{2 alpha D}{(1 + 4 alpha D t)^{3/2}} + frac{4 alpha^2 D x^2}{(1 + 4 alpha D t)^2} - k cdot frac{1}{sqrt{1 + 4 alpha D t}} right ) cdot frac{C_0}{A} e^{B} e^{-k t} ]Wait, but in the expression for ( D frac{partial^2 C}{partial x^2} - k C ), the coefficient is:[ D left( - frac{2 alpha}{1 + 4 alpha D t} + frac{4 alpha^2 x^2}{(1 + 4 alpha D t)^2} right ) - k ]Let me compute this:First term:[ - frac{2 alpha D}{1 + 4 alpha D t} ]Second term:[ frac{4 alpha^2 D x^2}{(1 + 4 alpha D t)^2} ]Third term:[ -k ]So, the total is:[ - frac{2 alpha D}{1 + 4 alpha D t} + frac{4 alpha^2 D x^2}{(1 + 4 alpha D t)^2} - k ]Compare this with the expression from ( frac{partial C}{partial t} ):[ - frac{2 alpha D}{(1 + 4 alpha D t)^{3/2}} + frac{4 alpha^2 D x^2}{(1 + 4 alpha D t)^2} - frac{k}{sqrt{1 + 4 alpha D t}} ]These are not the same. Therefore, my initial guess for the solution was incorrect.Hmm, so perhaps the solution is not just a simple Gaussian multiplied by an exponential decay. Maybe I need to go back to the Fourier transform approach and see where I went wrong.Earlier, when I did the Fourier transform, I ended up with an imaginary term in the exponent, which suggests that perhaps I need to consider the real part of the solution or that the solution involves both real and imaginary components, but since concentration is real, the imaginary parts must cancel out.Alternatively, perhaps I should have used a different method, such as separation of variables with a transformation.Let me try a substitution to simplify the PDE. Let me define a new function u(x,t) such that:[ C(x,t) = u(x,t) e^{-k t} ]Then, let's compute the derivatives:[ frac{partial C}{partial t} = frac{partial u}{partial t} e^{-k t} - k u e^{-k t} ][ frac{partial^2 C}{partial x^2} = frac{partial^2 u}{partial x^2} e^{-k t} ]Substitute into the PDE:[ frac{partial u}{partial t} e^{-k t} - k u e^{-k t} = D frac{partial^2 u}{partial x^2} e^{-k t} - k u e^{-k t} ]Simplify:Cancel out the ( -k u e^{-k t} ) terms on both sides:[ frac{partial u}{partial t} e^{-k t} = D frac{partial^2 u}{partial x^2} e^{-k t} ]Divide both sides by ( e^{-k t} ):[ frac{partial u}{partial t} = D frac{partial^2 u}{partial x^2} ]Ah, this is the standard heat equation! So, by making the substitution ( C = u e^{-k t} ), we've transformed the original PDE into the heat equation.Therefore, the solution for u(x,t) is the solution to the heat equation with initial condition:[ u(x,0) = C(x,0) e^{k cdot 0} = C_0 e^{-alpha x^2} ]So, the solution for u(x,t) is the convolution of the initial condition with the heat kernel.The heat kernel is:[ G(x,t) = frac{1}{sqrt{4 pi D t}} e^{ - frac{x^2}{4 D t} } ]Therefore, the solution for u(x,t) is:[ u(x,t) = int_{-infty}^{infty} G(x - x', t) C(x',0) dx' ]But since the initial condition is ( C(x,0) = C_0 e^{-alpha x^2} ), which is a Gaussian, the convolution of two Gaussians is another Gaussian.Specifically, the convolution of ( e^{-a x^2} ) and ( e^{-b x^2} ) is ( sqrt{frac{pi}{a + b}} e^{- frac{x^2}{4(a + b)} } ) or something similar. Let me recall the exact formula.The convolution of two Gaussians ( e^{-a x^2} ) and ( e^{-b x^2} ) is:[ int_{-infty}^{infty} e^{-a (x - x')^2} e^{-b x'^2} dx' = sqrt{frac{pi}{a + b}} e^{ - frac{a b x^2}{(a + b)} } ]Wait, no, more accurately, the convolution is:[ int_{-infty}^{infty} e^{-a (x - x')^2} e^{-b x'^2} dx' = sqrt{frac{pi}{a + b}} e^{ - frac{a b x^2}{(a + b)} } ]Wait, actually, let me compute it properly.Let me denote:[ int_{-infty}^{infty} e^{-a (x - x')^2} e^{-b x'^2} dx' ]Expand the exponent:[ -a (x^2 - 2 x x' + x'^2) - b x'^2 = -a x^2 + 2 a x x' - a x'^2 - b x'^2 = -a x^2 + 2 a x x' - (a + b) x'^2 ]Complete the square for x':[ - (a + b) x'^2 + 2 a x x' = - (a + b) left( x'^2 - frac{2 a x}{a + b} x' right ) ]Complete the square inside the parentheses:[ x'^2 - frac{2 a x}{a + b} x' = left( x' - frac{a x}{a + b} right )^2 - frac{a^2 x^2}{(a + b)^2} ]Therefore, the exponent becomes:[ -a x^2 - (a + b) left( left( x' - frac{a x}{a + b} right )^2 - frac{a^2 x^2}{(a + b)^2} right ) ]Simplify:[ -a x^2 - (a + b) left( x' - frac{a x}{a + b} right )^2 + frac{a^2 x^2}{a + b} ]Combine the x² terms:[ -a x^2 + frac{a^2 x^2}{a + b} = - frac{a (a + b) x^2 - a^2 x^2}{a + b} = - frac{a b x^2}{a + b} ]Therefore, the integral becomes:[ e^{ - frac{a b x^2}{a + b} } int_{-infty}^{infty} e^{ - (a + b) left( x' - frac{a x}{a + b} right )^2 } dx' ]The integral is a Gaussian integral:[ int_{-infty}^{infty} e^{-c y^2} dy = sqrt{frac{pi}{c}} ]Here, c = a + b, so the integral is ( sqrt{frac{pi}{a + b}} )Therefore, the convolution is:[ sqrt{frac{pi}{a + b}} e^{ - frac{a b x^2}{a + b} } ]So, in our case, the heat kernel is ( G(x,t) = frac{1}{sqrt{4 pi D t}} e^{ - frac{x^2}{4 D t} } ), which corresponds to a = 1/(4 D t), and the initial condition is ( C(x,0) = C_0 e^{-alpha x^2} ), which corresponds to b = α.Therefore, the convolution integral becomes:[ u(x,t) = C_0 sqrt{frac{pi}{alpha + 1/(4 D t)}} e^{ - frac{alpha cdot 1/(4 D t) x^2}{alpha + 1/(4 D t)} } ]Simplify the denominator:[ alpha + frac{1}{4 D t} = frac{4 D t alpha + 1}{4 D t} ]So,[ sqrt{frac{pi}{alpha + 1/(4 D t)}} = sqrt{frac{pi cdot 4 D t}{4 D t alpha + 1}} = 2 sqrt{frac{pi D t}{4 D t alpha + 1}} ]And the exponent:[ - frac{alpha cdot 1/(4 D t) x^2}{alpha + 1/(4 D t)} = - frac{alpha x^2}{4 D t alpha + 1} ]Therefore, u(x,t) is:[ u(x,t) = C_0 cdot 2 sqrt{frac{pi D t}{4 D t alpha + 1}} e^{ - frac{alpha x^2}{4 D t alpha + 1} } ]But wait, the convolution integral should give us u(x,t) as:[ u(x,t) = int_{-infty}^{infty} G(x - x', t) C(x',0) dx' = C_0 sqrt{frac{pi}{alpha + 1/(4 D t)}} e^{ - frac{alpha x^2}{4 D t alpha + 1} } ]Wait, but I think I missed a factor. Let me double-check.The convolution formula gives:[ u(x,t) = sqrt{frac{pi}{alpha + 1/(4 D t)}} e^{ - frac{alpha cdot 1/(4 D t) x^2}{alpha + 1/(4 D t)} } ]But since the initial condition is scaled by C_0, it should be:[ u(x,t) = C_0 sqrt{frac{pi}{alpha + 1/(4 D t)}} e^{ - frac{alpha x^2}{4 D t alpha + 1} } ]Wait, but the heat kernel is ( frac{1}{sqrt{4 pi D t}} e^{-x^2/(4 D t)} ), so when convolving with ( C_0 e^{-alpha x^2} ), the result is:[ u(x,t) = C_0 cdot sqrt{frac{pi}{alpha + 1/(4 D t)}} e^{ - frac{alpha x^2}{4 D t alpha + 1} } ]Therefore, the solution for u(x,t) is:[ u(x,t) = C_0 sqrt{frac{pi}{alpha + 1/(4 D t)}} e^{ - frac{alpha x^2}{4 D t alpha + 1} } ]But wait, the units don't seem to match. Let me check:The initial condition is C_0 e^{-α x²}, so u(x,0) = C_0 e^{-α x²}. But according to the solution, at t=0, we have:[ u(x,0) = C_0 sqrt{frac{pi}{alpha + infty}} e^{ - frac{alpha x^2}{infty} } ]Wait, that doesn't make sense. As t approaches 0, 1/(4 D t) approaches infinity, so the denominator in the square root becomes dominated by 1/(4 D t), so:[ sqrt{frac{pi}{alpha + 1/(4 D t)}} approx sqrt{frac{pi cdot 4 D t}{1}} = 2 sqrt{pi D t} ]And the exponent becomes:[ - frac{alpha x^2}{1/(4 D t)} = -4 D t alpha x^2 ]So, as t approaches 0, u(x,t) approaches:[ C_0 cdot 2 sqrt{pi D t} e^{-4 D t alpha x^2} ]But the initial condition is u(x,0) = C_0 e^{-α x²}, so this doesn't match. Therefore, there must be a mistake in the convolution.Wait, perhaps I should have considered that the heat kernel is ( frac{1}{sqrt{4 pi D t}} e^{-x^2/(4 D t)} ), so when convolving with ( C_0 e^{-α x²} ), the result is:[ u(x,t) = C_0 int_{-infty}^{infty} frac{1}{sqrt{4 pi D t}} e^{- (x - x')²/(4 D t)} e^{- α x'²} dx' ]Which is:[ C_0 cdot sqrt{frac{pi}{α + 1/(4 D t)}} e^{- frac{α x²}{4 D t α + 1} } ]Wait, but at t=0, this becomes:[ C_0 cdot sqrt{frac{pi}{α + infty}} e^{-0} = C_0 cdot 0 cdot 1 = 0 ]Which is not correct because u(x,0) should be C_0 e^{-α x²}. Therefore, my convolution approach must be incorrect.Wait, perhaps I need to consider that the heat equation solution is:[ u(x,t) = frac{1}{sqrt{4 pi D t}} int_{-infty}^{infty} e^{- (x - x')²/(4 D t)} C(x',0) dx' ]So, substituting C(x',0) = C_0 e^{-α x'²}:[ u(x,t) = C_0 frac{1}{sqrt{4 pi D t}} int_{-infty}^{infty} e^{- (x - x')²/(4 D t)} e^{- α x'²} dx' ]This integral is the same as the convolution of two Gaussians, which we computed earlier as:[ sqrt{frac{pi}{α + 1/(4 D t)}} e^{- frac{α x²}{4 D t α + 1} } ]But then, u(x,t) is:[ u(x,t) = C_0 frac{1}{sqrt{4 pi D t}} cdot sqrt{frac{pi}{α + 1/(4 D t)}} e^{- frac{α x²}{4 D t α + 1} } ]Simplify:[ u(x,t) = C_0 frac{1}{sqrt{4 pi D t}} cdot sqrt{frac{pi}{α + 1/(4 D t)}} e^{- frac{α x²}{4 D t α + 1} } ]Simplify the constants:[ frac{1}{sqrt{4 pi D t}} cdot sqrt{frac{pi}{α + 1/(4 D t)}} = frac{1}{sqrt{4 D t}} cdot frac{1}{sqrt{α + 1/(4 D t)}} ]Multiply numerator and denominator by sqrt(4 D t):[ frac{1}{sqrt{4 D t} cdot sqrt{α + 1/(4 D t)}} = frac{1}{sqrt{4 D t α + 1}} ]Therefore, u(x,t) simplifies to:[ u(x,t) = frac{C_0}{sqrt{4 D t α + 1}} e^{- frac{α x²}{4 D t α + 1} } ]Ah, that makes more sense. Now, let's check the initial condition:At t=0, 4 D t α approaches 0, so:[ u(x,0) = frac{C_0}{sqrt{0 + 1}} e^{- frac{α x²}{0 + 1} } = C_0 e^{-α x²} ]Which matches the initial condition. Great!Therefore, the solution for u(x,t) is:[ u(x,t) = frac{C_0}{sqrt{1 + 4 D α t}} e^{- frac{α x²}{1 + 4 D α t} } ]Since C(x,t) = u(x,t) e^{-k t}, the solution for C(x,t) is:[ C(x,t) = frac{C_0}{sqrt{1 + 4 D α t}} e^{- frac{α x²}{1 + 4 D α t} } e^{-k t} ]Alternatively, combining the exponentials:[ C(x,t) = frac{C_0}{sqrt{1 + 4 D α t}} e^{- left( frac{α x²}{1 + 4 D α t} + k t right ) } ]But usually, it's written as a product of the Gaussian and the exponential decay.So, the general solution is:[ C(x,t) = frac{C_0}{sqrt{1 + 4 D α t}} e^{- frac{α x²}{1 + 4 D α t} } e^{-k t} ]This satisfies both the PDE and the initial condition.Now, moving on to part 2: finding the time T at which the concentration at position x = L drops below C_threshold.So, we need to solve for T in:[ C(L, T) = C_{text{threshold}} ]Substitute the expression for C(x,t):[ frac{C_0}{sqrt{1 + 4 D α T}} e^{- frac{α L²}{1 + 4 D α T} } e^{-k T} = C_{text{threshold}} ]This equation is transcendental and likely doesn't have a closed-form solution, but we can express T implicitly or solve it numerically. However, since the question asks to express T in terms of the given parameters, we might need to manipulate the equation to isolate T.Let me denote:Let’s set ( S = 1 + 4 D α T ). Then, the equation becomes:[ frac{C_0}{sqrt{S}} e^{- frac{α L²}{S} } e^{-k T} = C_{text{threshold}} ]But since S = 1 + 4 D α T, we can express T as:[ T = frac{S - 1}{4 D α} ]Substitute back into the equation:[ frac{C_0}{sqrt{S}} e^{- frac{α L²}{S} } e^{-k cdot frac{S - 1}{4 D α} } = C_{text{threshold}} ]This equation is still implicit in S, so solving for S (and hence T) would require numerical methods. However, perhaps we can make an approximation or rearrange terms.Alternatively, take natural logarithm on both sides:[ ln left( frac{C_0}{sqrt{S}} e^{- frac{α L²}{S} } e^{-k T} right ) = ln C_{text{threshold}} ]Simplify the left side:[ ln C_0 - frac{1}{2} ln S - frac{α L²}{S} - k T = ln C_{text{threshold}} ]Substitute T = (S - 1)/(4 D α):[ ln C_0 - frac{1}{2} ln S - frac{α L²}{S} - k cdot frac{S - 1}{4 D α} = ln C_{text{threshold}} ]This is a complicated equation in S. It might not be possible to solve analytically, so perhaps we can leave the answer in terms of S or express T implicitly.Alternatively, if we assume that 4 D α T is much larger than 1, i.e., S ≈ 4 D α T, then we can approximate:[ C(x,t) ≈ frac{C_0}{sqrt{4 D α T}} e^{- frac{α x²}{4 D α T} } e^{-k T} = frac{C_0}{2 sqrt{D α T}} e^{- frac{x²}{4 D T} } e^{-k T} ]But this is only valid for large T, which might not be the case here.Alternatively, if k is small compared to the other terms, we might neglect it, but since k is a metabolism rate, it might not be negligible.Given that, perhaps the best we can do is express T implicitly. However, the problem asks to express T in terms of the given parameters, so maybe we can write it as:[ frac{C_0}{sqrt{1 + 4 D α T}} e^{- frac{α L²}{1 + 4 D α T} - k T} = C_{text{threshold}} ]Taking natural logarithm:[ ln C_0 - frac{1}{2} ln(1 + 4 D α T) - frac{α L²}{1 + 4 D α T} - k T = ln C_{text{threshold}} ]This is the implicit equation for T. Therefore, T cannot be expressed in a simple closed-form expression and would require numerical methods to solve.However, perhaps we can make an approximation or rearrange terms to express T in terms of the other variables. Alternatively, if we assume that the term ( frac{α L²}{1 + 4 D α T} ) is small, but that might not hold.Alternatively, let me consider that the equation is:[ frac{C_0}{sqrt{1 + 4 D α T}} e^{- frac{α L²}{1 + 4 D α T} - k T} = C_{text{threshold}} ]Let me denote ( u = 1 + 4 D α T ), then T = (u - 1)/(4 D α)Substitute into the equation:[ frac{C_0}{sqrt{u}} e^{- frac{α L²}{u} - k cdot frac{u - 1}{4 D α} } = C_{text{threshold}} ]This is still implicit in u.Alternatively, perhaps we can write:[ frac{C_0}{sqrt{u}} e^{- frac{α L²}{u} } e^{- frac{k (u - 1)}{4 D α} } = C_{text{threshold}} ]But I don't see a straightforward way to solve for u.Therefore, the answer for part 2 is that T is the solution to the equation:[ frac{C_0}{sqrt{1 + 4 D α T}} e^{- frac{α L²}{1 + 4 D α T} - k T} = C_{text{threshold}} ]Which can be written as:[ frac{C_0}{sqrt{1 + 4 D α T}} e^{- left( frac{α L²}{1 + 4 D α T} + k T right ) } = C_{text{threshold}} ]This is the implicit equation for T, which would need to be solved numerically given the values of D, k, C_0, α, L, and C_threshold.Alternatively, if we take logarithms:[ ln C_0 - frac{1}{2} ln(1 + 4 D α T) - frac{α L²}{1 + 4 D α T} - k T = ln C_{text{threshold}} ]This is the equation that T must satisfy. Therefore, T is given implicitly by this equation.So, summarizing:1. The general solution is:[ C(x,t) = frac{C_0}{sqrt{1 + 4 D α t}} e^{- frac{α x²}{1 + 4 D α t} } e^{-k t} ]2. The time T is the solution to:[ frac{C_0}{sqrt{1 + 4 D α T}} e^{- frac{α L²}{1 + 4 D α T} - k T} = C_{text{threshold}} ]Which can be written as:[ frac{C_0}{sqrt{1 + 4 D α T}} e^{- left( frac{α L²}{1 + 4 D α T} + k T right ) } = C_{text{threshold}} ]Therefore, T is expressed implicitly in terms of the given parameters.</think>

answer:<think>Alright, so I've got this problem about modeling the population dynamics of two regions during the Russian Revolution using differential equations. It's split into two parts: first, solving the system of differential equations, and second, analyzing the equilibrium points and their stability. Let me try to tackle this step by step.Starting with part 1: solving the system. The equations given are:For region A:[ frac{dP_A}{dt} = k_1 P_A left(1 - frac{P_A}{K_A}right) - m P_A P_B ]For region B:[ frac{dP_B}{dt} = k_2 P_B left(1 - frac{P_B}{K_B}right) + m P_A P_B - d P_B ]These are coupled nonlinear differential equations because each equation involves terms from both ( P_A ) and ( P_B ). Nonlinear systems can be tricky because they often don't have closed-form solutions, but let me see if I can manipulate them or find some substitution that simplifies things.First, let me rewrite the equations for clarity:1. ( frac{dP_A}{dt} = k_1 P_A - frac{k_1}{K_A} P_A^2 - m P_A P_B )2. ( frac{dP_B}{dt} = k_2 P_B - frac{k_2}{K_B} P_B^2 + m P_A P_B - d P_B )Hmm, maybe I can combine these equations somehow. Let me try adding them together to see if something cancels out.Adding equation 1 and equation 2:( frac{dP_A}{dt} + frac{dP_B}{dt} = (k_1 P_A - frac{k_1}{K_A} P_A^2 - m P_A P_B) + (k_2 P_B - frac{k_2}{K_B} P_B^2 + m P_A P_B - d P_B) )Simplify the right-hand side:- The ( -m P_A P_B ) and ( +m P_A P_B ) terms cancel out.- Combine the ( P_A ) and ( P_B ) terms: ( k_1 P_A + (k_2 - d) P_B )- The quadratic terms remain: ( -frac{k_1}{K_A} P_A^2 - frac{k_2}{K_B} P_B^2 )So, the sum becomes:( frac{d}{dt}(P_A + P_B) = k_1 P_A + (k_2 - d) P_B - frac{k_1}{K_A} P_A^2 - frac{k_2}{K_B} P_B^2 )Hmm, not sure if that helps directly, but it's an interesting observation. Maybe instead of adding, I can subtract them? Let's see.Subtracting equation 2 from equation 1:( frac{dP_A}{dt} - frac{dP_B}{dt} = (k_1 P_A - frac{k_1}{K_A} P_A^2 - m P_A P_B) - (k_2 P_B - frac{k_2}{K_B} P_B^2 + m P_A P_B - d P_B) )Simplify:- ( k_1 P_A - frac{k_1}{K_A} P_A^2 - m P_A P_B - k_2 P_B + frac{k_2}{K_B} P_B^2 - m P_A P_B + d P_B )- Combine like terms: ( k_1 P_A - frac{k_1}{K_A} P_A^2 - 2m P_A P_B + (-k_2 + d) P_B + frac{k_2}{K_B} P_B^2 )This also seems complicated. Maybe another approach is needed.Alternatively, perhaps I can express one variable in terms of the other. Let me see if I can solve for ( frac{dP_A}{dt} ) and ( frac{dP_B}{dt} ) in terms of each other.From equation 1:( frac{dP_A}{dt} = k_1 P_A left(1 - frac{P_A}{K_A}right) - m P_A P_B )Let me factor out ( P_A ):( frac{dP_A}{dt} = P_A left( k_1 left(1 - frac{P_A}{K_A}right) - m P_B right) )Similarly, equation 2:( frac{dP_B}{dt} = k_2 P_B left(1 - frac{P_B}{K_B}right) + m P_A P_B - d P_B )Factor out ( P_B ):( frac{dP_B}{dt} = P_B left( k_2 left(1 - frac{P_B}{K_B}right) + m P_A - d right) )Hmm, so each equation is of the form ( frac{dP}{dt} = P (f(P) + g(Q)) ), where ( f ) is a function of the same variable and ( g ) is a function of the other variable.This structure suggests that it might be a Lotka-Volterra type system, but with some modifications. In the classic Lotka-Volterra model, you have one species growing logistically and the other having a term dependent on the first. Here, both regions have logistic growth terms and a coupling term through migration or interaction.Given that these are coupled logistic equations, finding an explicit solution might be difficult. I recall that for Lotka-Volterra systems, solutions can sometimes be found in terms of integrals, but they often don't have closed-form expressions. So, perhaps I need to consider if there's a substitution or transformation that can linearize the system or make it more manageable.Alternatively, maybe I can look for equilibrium points first, as that might help in understanding the behavior of the system, which is actually part 2 of the problem. But since part 1 asks for solving the system, I need to see if it's possible.Another thought: if the coupling term ( m P_A P_B ) is small or if ( m ) is zero, the equations decouple into two logistic equations, which have known solutions. But since ( m ) is a migration factor influenced by political instability, it's likely non-zero, so the coupling is significant.Wait, perhaps I can consider a change of variables to simplify the equations. Let me define ( x = P_A ) and ( y = P_B ). Then, the system becomes:1. ( frac{dx}{dt} = k_1 x (1 - frac{x}{K_A}) - m x y )2. ( frac{dy}{dt} = k_2 y (1 - frac{y}{K_B}) + m x y - d y )This doesn't immediately suggest a substitution, but maybe I can divide the two equations to get a relationship between ( frac{dy}{dx} ) and ( x ) and ( y ).Dividing equation 2 by equation 1:( frac{dy}{dx} = frac{k_2 y (1 - frac{y}{K_B}) + m x y - d y}{k_1 x (1 - frac{x}{K_A}) - m x y} )This is a first-order ordinary differential equation in terms of ( y ) and ( x ), but it's still nonlinear and doesn't look easily integrable. Maybe I can rearrange terms or factor something out.Looking at the numerator and denominator:Numerator: ( y [k_2 (1 - frac{y}{K_B}) + m x - d] )Denominator: ( x [k_1 (1 - frac{x}{K_A}) - m y] )So,( frac{dy}{dx} = frac{y [k_2 (1 - frac{y}{K_B}) + m x - d]}{x [k_1 (1 - frac{x}{K_A}) - m y]} )This still seems complicated, but perhaps I can separate variables or make a substitution. Let me try to see if this can be expressed as a Bernoulli equation or something similar.Alternatively, maybe I can consider the system in terms of dimensionless variables. Let me define:( u = frac{x}{K_A} ), ( v = frac{y}{K_B} ), and ( tau = k_1 t )Then, ( x = K_A u ), ( y = K_B v ), and ( t = tau / k_1 )Compute the derivatives:( frac{dx}{dt} = K_A frac{du}{dtau} cdot frac{dtau}{dt} = K_A k_1 frac{du}{dtau} )Similarly, ( frac{dy}{dt} = K_B k_1 frac{dv}{dtau} )Substitute into the original equations:1. ( K_A k_1 frac{du}{dtau} = k_1 K_A u (1 - u) - m K_A u K_B v ) Simplify: ( frac{du}{dtau} = u (1 - u) - frac{m K_B}{k_1} u v )2. ( K_B k_1 frac{dv}{dtau} = k_2 K_B v (1 - v) + m K_A u K_B v - d K_B v ) Simplify: ( frac{dv}{dtau} = frac{k_2}{k_1} v (1 - v) + frac{m K_A}{k_1} u v - frac{d}{k_1} v )Let me define some dimensionless parameters to simplify further:Let ( alpha = frac{m K_B}{k_1} ), ( beta = frac{k_2}{k_1} ), ( gamma = frac{m K_A}{k_1} ), and ( delta = frac{d}{k_1} )Then, the system becomes:1. ( frac{du}{dtau} = u (1 - u) - alpha u v )2. ( frac{dv}{dtau} = beta v (1 - v) + gamma u v - delta v )This is still a nonlinear system, but perhaps in terms of these dimensionless variables, it's easier to analyze. However, solving it analytically might still be challenging.Another approach: maybe assume that one variable can be expressed as a function of the other. For example, suppose we express ( v ) in terms of ( u ). But without knowing the functional form, that might not help.Alternatively, consider if the system can be transformed into a linear system through some substitution, but given the quadratic terms, that seems unlikely.Wait, perhaps I can look for invariant curves or conserved quantities. If there's a function ( H(u, v) ) such that ( frac{dH}{dtau} = 0 ), then ( H(u, v) ) is constant along the trajectories. Let me try to find such a function.Compute ( frac{dH}{dtau} = frac{partial H}{partial u} frac{du}{dtau} + frac{partial H}{partial v} frac{dv}{dtau} )Set this equal to zero:( frac{partial H}{partial u} [u (1 - u) - alpha u v] + frac{partial H}{partial v} [beta v (1 - v) + gamma u v - delta v] = 0 )This is a PDE for ( H(u, v) ). Solving this might be non-trivial, but perhaps we can guess a form for ( H ). For example, suppose ( H ) is a combination of ( u ) and ( v ) terms. Maybe something like ( H = A u + B v + C u v ). Let me try that.Compute the partial derivatives:( frac{partial H}{partial u} = A + C v )( frac{partial H}{partial v} = B + C u )Substitute into the PDE:( (A + C v)[u (1 - u) - alpha u v] + (B + C u)[beta v (1 - v) + gamma u v - delta v] = 0 )This must hold for all ( u ) and ( v ), so we can equate coefficients of like terms to zero.Let me expand the terms:First term:( (A + C v)(u - u^2 - alpha u v) = A u - A u^2 - A alpha u v + C v u - C v u^2 - C alpha v^2 u )Second term:( (B + C u)(beta v - beta v^2 + gamma u v - delta v) )Let me expand this:First, group the terms inside the second parenthesis:( beta v - beta v^2 + gamma u v - delta v = (beta - delta) v - beta v^2 + gamma u v )Now, multiply by ( (B + C u) ):( B [(beta - delta) v - beta v^2 + gamma u v] + C u [(beta - delta) v - beta v^2 + gamma u v] )Expanding:( B (beta - delta) v - B beta v^2 + B gamma u v + C (beta - delta) u v - C beta u v^2 + C gamma u^2 v )Now, combine all terms from both expansions:From first term:1. ( A u )2. ( -A u^2 )3. ( -A alpha u v )4. ( C u v )5. ( -C v u^2 )6. ( -C alpha v^2 u )From second term:7. ( B (beta - delta) v )8. ( -B beta v^2 )9. ( B gamma u v )10. ( C (beta - delta) u v )11. ( -C beta u v^2 )12. ( C gamma u^2 v )Now, collect like terms:- Terms with ( u ): 1. ( A u )- Terms with ( v ): 7. ( B (beta - delta) v )- Terms with ( u^2 ): 2. ( -A u^2 )- Terms with ( v^2 ): 8. ( -B beta v^2 )- Terms with ( u v ): 3. ( -A alpha u v ), 4. ( C u v ), 9. ( B gamma u v ), 10. ( C (beta - delta) u v )- Terms with ( u^2 v ): 5. ( -C v u^2 ), 12. ( C gamma u^2 v )- Terms with ( u v^2 ): 6. ( -C alpha v^2 u ), 11. ( -C beta u v^2 )Now, set the coefficients of each term to zero:1. Coefficient of ( u ): ( A = 0 )2. Coefficient of ( v ): ( B (beta - delta) = 0 )3. Coefficient of ( u^2 ): ( -A = 0 ) (already satisfied since A=0)4. Coefficient of ( v^2 ): ( -B beta = 0 )5. Coefficient of ( u v ): ( -A alpha + C + B gamma + C (beta - delta) = 0 ) Since A=0, this simplifies to ( C + B gamma + C (beta - delta) = 0 )6. Coefficient of ( u^2 v ): ( -C + C gamma = 0 ) → ( C ( -1 + gamma ) = 0 )7. Coefficient of ( u v^2 ): ( -C alpha - C beta = 0 ) → ( -C (alpha + beta) = 0 )Now, let's solve these equations step by step.From equation 2: ( B (beta - delta) = 0 )From equation 4: ( -B beta = 0 )Assuming ( beta neq 0 ) (since ( k_2 ) is a growth rate and likely positive), equation 4 implies ( B = 0 ).If ( B = 0 ), then equation 2 is satisfied regardless of ( beta - delta ).From equation 6: ( C ( -1 + gamma ) = 0 )From equation 7: ( -C (alpha + beta) = 0 )Assuming ( alpha + beta neq 0 ), equation 7 implies ( C = 0 ). Then, equation 6 is satisfied.But if ( C = 0 ), then from equation 5: ( 0 + 0 + 0 = 0 ), which is satisfied.So, the only solution is ( A = 0 ), ( B = 0 ), ( C = 0 ), which means ( H(u, v) ) is a constant, which doesn't help us find a non-trivial conserved quantity.Therefore, this approach doesn't yield a useful invariant. Maybe another form for ( H ) is needed, but this might be too time-consuming and not fruitful.Given that analytical solutions are proving difficult, perhaps I should consider that the problem expects an analysis rather than an explicit solution. Wait, part 1 says "solve the system of differential equations," but given the nonlinearity, it's unlikely to have a closed-form solution. Maybe the problem expects us to set up the equations and recognize that they form a coupled logistic system, and perhaps find equilibrium points and their stability, which is part 2.Alternatively, perhaps the problem is expecting to linearize the system around equilibrium points and analyze stability, which would be part 2. But part 1 specifically says "solve the system," so maybe I need to proceed differently.Wait, another thought: if I assume that the populations reach equilibrium quickly, then perhaps I can find the equilibrium points and describe the behavior around them, but that's more part 2.Alternatively, maybe I can consider perturbations around equilibrium points, but again, that's part 2.Hmm, perhaps the problem is expecting us to recognize that this is a system of coupled logistic equations and that solving it analytically is not straightforward, so instead, we can discuss the methods to approach it, such as numerical solutions or phase plane analysis.But the question says "solve the system," so maybe I need to proceed with finding equilibrium points and analyzing their stability, which would be part 2, but perhaps the solution to part 1 is just acknowledging that an explicit solution is difficult and instead focusing on equilibrium analysis.Alternatively, maybe I can find an integrating factor or some other technique, but I don't see an obvious path.Wait, perhaps I can consider the case where ( m = 0 ), which decouples the equations. Then, each region follows a logistic growth:For region A: ( frac{dP_A}{dt} = k_1 P_A (1 - frac{P_A}{K_A}) )Solution: ( P_A(t) = frac{K_A P_{A0}}{P_{A0} + (K_A - P_{A0}) e^{-k_1 t}} )Similarly, for region B: ( frac{dP_B}{dt} = k_2 P_B (1 - frac{P_B}{K_B}) - d P_B )This is a logistic equation with a death term. Let me rewrite it:( frac{dP_B}{dt} = (k_2 - d) P_B (1 - frac{P_B}{K_B}) - d P_B frac{P_B}{K_B} )Wait, no, actually:Wait, ( frac{dP_B}{dt} = k_2 P_B (1 - frac{P_B}{K_B}) - d P_B )Factor out ( P_B ):( frac{dP_B}{dt} = P_B [k_2 (1 - frac{P_B}{K_B}) - d] )Let me define ( k_2' = k_2 - d ), then:( frac{dP_B}{dt} = k_2' P_B (1 - frac{P_B}{K_B}) + d P_B ( - frac{P_B}{K_B} ) )Wait, no, that's not correct. Let me re-express:( frac{dP_B}{dt} = k_2 P_B - frac{k_2}{K_B} P_B^2 - d P_B )Combine like terms:( frac{dP_B}{dt} = (k_2 - d) P_B - frac{k_2}{K_B} P_B^2 )This is a logistic equation with growth rate ( k_2' = k_2 - d ) and carrying capacity ( K_B' = frac{(k_2 - d) K_B}{k_2} ), provided ( k_2 > d ). If ( k_2 leq d ), the population will decline to zero.So, the solution for region B when ( m = 0 ) is:( P_B(t) = frac{K_B' P_{B0}}{P_{B0} + (K_B' - P_{B0}) e^{-k_2' t}} ), where ( K_B' = frac{(k_2 - d) K_B}{k_2} ) if ( k_2 > d ).But in the original problem, ( m ) is not zero, so the coupling complicates things. Therefore, without knowing specific values for the parameters, it's difficult to provide an explicit solution. Hence, perhaps the answer to part 1 is that an explicit solution is not feasible analytically and that numerical methods or equilibrium analysis is required.Moving on to part 2: Analyze the equilibrium points and their stability.Equilibrium points occur where ( frac{dP_A}{dt} = 0 ) and ( frac{dP_B}{dt} = 0 ).So, set the derivatives equal to zero:1. ( k_1 P_A (1 - frac{P_A}{K_A}) - m P_A P_B = 0 )2. ( k_2 P_B (1 - frac{P_B}{K_B}) + m P_A P_B - d P_B = 0 )Let me factor these equations:From equation 1:( P_A [k_1 (1 - frac{P_A}{K_A}) - m P_B] = 0 )So, either ( P_A = 0 ) or ( k_1 (1 - frac{P_A}{K_A}) - m P_B = 0 )From equation 2:( P_B [k_2 (1 - frac{P_B}{K_B}) + m P_A - d] = 0 )So, either ( P_B = 0 ) or ( k_2 (1 - frac{P_B}{K_B}) + m P_A - d = 0 )Now, let's find all possible equilibrium points by considering combinations of these solutions.Case 1: ( P_A = 0 ) and ( P_B = 0 )This is the trivial equilibrium where both populations are extinct.Case 2: ( P_A = 0 ), ( P_B neq 0 )From equation 1, ( P_A = 0 ). Substitute into equation 2:( k_2 P_B (1 - frac{P_B}{K_B}) - d P_B = 0 )Factor out ( P_B ):( P_B [k_2 (1 - frac{P_B}{K_B}) - d] = 0 )So, either ( P_B = 0 ) (which is case 1) or ( k_2 (1 - frac{P_B}{K_B}) - d = 0 )Solving for ( P_B ):( k_2 (1 - frac{P_B}{K_B}) = d )( 1 - frac{P_B}{K_B} = frac{d}{k_2} )( frac{P_B}{K_B} = 1 - frac{d}{k_2} )( P_B = K_B (1 - frac{d}{k_2}) )This is valid only if ( 1 - frac{d}{k_2} geq 0 ), i.e., ( k_2 geq d ). So, equilibrium point is ( (0, K_B (1 - frac{d}{k_2})) ) if ( k_2 geq d ).Case 3: ( P_B = 0 ), ( P_A neq 0 )From equation 2, ( P_B = 0 ). Substitute into equation 1:( k_1 P_A (1 - frac{P_A}{K_A}) = 0 )So, either ( P_A = 0 ) (case 1) or ( 1 - frac{P_A}{K_A} = 0 ) → ( P_A = K_A )Thus, equilibrium point is ( (K_A, 0) )Case 4: ( P_A neq 0 ) and ( P_B neq 0 )From equation 1:( k_1 (1 - frac{P_A}{K_A}) - m P_B = 0 ) → ( m P_B = k_1 (1 - frac{P_A}{K_A}) ) → ( P_B = frac{k_1}{m} (1 - frac{P_A}{K_A}) )From equation 2:( k_2 (1 - frac{P_B}{K_B}) + m P_A - d = 0 )Substitute ( P_B ) from equation 1 into equation 2:( k_2 left(1 - frac{1}{K_B} cdot frac{k_1}{m} (1 - frac{P_A}{K_A}) right) + m P_A - d = 0 )Simplify:( k_2 - frac{k_2 k_1}{m K_B} (1 - frac{P_A}{K_A}) + m P_A - d = 0 )Let me rearrange terms:( - frac{k_2 k_1}{m K_B} (1 - frac{P_A}{K_A}) + m P_A + (k_2 - d) = 0 )Multiply through by ( m K_B ) to eliminate denominators:( -k_2 k_1 (1 - frac{P_A}{K_A}) + m^2 K_B P_A + (k_2 - d) m K_B = 0 )Expand the first term:( -k_2 k_1 + frac{k_2 k_1}{K_A} P_A + m^2 K_B P_A + (k_2 - d) m K_B = 0 )Combine like terms:( left( frac{k_2 k_1}{K_A} + m^2 K_B right) P_A + (-k_2 k_1 + (k_2 - d) m K_B) = 0 )Solve for ( P_A ):( P_A = frac{k_2 k_1 - (k_2 - d) m K_B}{frac{k_2 k_1}{K_A} + m^2 K_B} )Let me factor out ( k_2 ) from the numerator:( P_A = frac{k_2 (k_1 - m K_B) + d m K_B}{frac{k_2 k_1}{K_A} + m^2 K_B} )Alternatively, write it as:( P_A = frac{k_2 k_1 - k_2 m K_B + d m K_B}{frac{k_2 k_1}{K_A} + m^2 K_B} )Factor numerator:( P_A = frac{k_2 (k_1 - m K_B) + d m K_B}{frac{k_2 k_1}{K_A} + m^2 K_B} )Similarly, once ( P_A ) is found, substitute back into ( P_B = frac{k_1}{m} (1 - frac{P_A}{K_A}) ) to find ( P_B ).This gives the non-trivial equilibrium point ( (P_A^*, P_B^*) ) where both populations are positive.Now, to determine the stability of these equilibrium points, we need to linearize the system around each equilibrium and analyze the eigenvalues of the Jacobian matrix.The Jacobian matrix ( J ) is given by:[ J = begin{bmatrix}frac{partial}{partial P_A} left( k_1 P_A (1 - frac{P_A}{K_A}) - m P_A P_B right) & frac{partial}{partial P_B} left( k_1 P_A (1 - frac{P_A}{K_A}) - m P_A P_B right) frac{partial}{partial P_A} left( k_2 P_B (1 - frac{P_B}{K_B}) + m P_A P_B - d P_B right) & frac{partial}{partial P_B} left( k_2 P_B (1 - frac{P_B}{K_B}) + m P_A P_B - d P_B right)end{bmatrix} ]Compute each partial derivative:First row, first column:( frac{partial}{partial P_A} [k_1 P_A (1 - frac{P_A}{K_A}) - m P_A P_B] = k_1 (1 - frac{P_A}{K_A}) - k_1 frac{P_A}{K_A} - m P_B = k_1 (1 - frac{2 P_A}{K_A}) - m P_B )First row, second column:( frac{partial}{partial P_B} [k_1 P_A (1 - frac{P_A}{K_A}) - m P_A P_B] = -m P_A )Second row, first column:( frac{partial}{partial P_A} [k_2 P_B (1 - frac{P_B}{K_B}) + m P_A P_B - d P_B] = m P_B )Second row, second column:( frac{partial}{partial P_B} [k_2 P_B (1 - frac{P_B}{K_B}) + m P_A P_B - d P_B] = k_2 (1 - frac{P_B}{K_B}) - frac{k_2 P_B}{K_B} + m P_A - d = k_2 (1 - frac{2 P_B}{K_B}) + m P_A - d )So, the Jacobian matrix is:[ J = begin{bmatrix}k_1 (1 - frac{2 P_A}{K_A}) - m P_B & -m P_A m P_B & k_2 (1 - frac{2 P_B}{K_B}) + m P_A - dend{bmatrix} ]Now, evaluate ( J ) at each equilibrium point.1. Trivial equilibrium ( (0, 0) ):[ J(0,0) = begin{bmatrix}k_1 & 0 0 & -dend{bmatrix} ]The eigenvalues are ( k_1 ) and ( -d ). Since ( k_1 > 0 ) and ( d > 0 ), this equilibrium is a saddle point. Therefore, it's unstable.2. Equilibrium ( (0, P_B^*) ) where ( P_B^* = K_B (1 - frac{d}{k_2}) ) (assuming ( k_2 > d )):Evaluate ( J ) at ( (0, P_B^*) ):First, compute the partial derivatives:- ( frac{partial}{partial P_A} ) term: ( k_1 (1 - 0) - m P_B^* = k_1 - m P_B^* )- ( frac{partial}{partial P_B} ) term: ( -m cdot 0 = 0 )- ( frac{partial}{partial P_A} ) term for second row: ( m P_B^* )- ( frac{partial}{partial P_B} ) term: ( k_2 (1 - frac{2 P_B^*}{K_B}) + 0 - d )Compute each:First row, first column: ( k_1 - m P_B^* = k_1 - m K_B (1 - frac{d}{k_2}) )First row, second column: 0Second row, first column: ( m P_B^* = m K_B (1 - frac{d}{k_2}) )Second row, second column: ( k_2 (1 - frac{2 P_B^*}{K_B}) - d = k_2 (1 - 2 (1 - frac{d}{k_2})) - d = k_2 (1 - 2 + frac{2d}{k_2}) - d = k_2 (-1 + frac{2d}{k_2}) - d = -k_2 + 2d - d = -k_2 + d )So, the Jacobian is:[ J(0, P_B^*) = begin{bmatrix}k_1 - m K_B (1 - frac{d}{k_2}) & 0 m K_B (1 - frac{d}{k_2}) & -k_2 + dend{bmatrix} ]The eigenvalues are the diagonal elements since it's a triangular matrix.Eigenvalues:- ( lambda_1 = k_1 - m K_B (1 - frac{d}{k_2}) )- ( lambda_2 = -k_2 + d )For stability, both eigenvalues must have negative real parts.So, conditions:1. ( k_1 - m K_B (1 - frac{d}{k_2}) < 0 )2. ( -k_2 + d < 0 ) → ( d < k_2 )But in this case, ( P_B^* ) exists only if ( k_2 > d ), so condition 2 is satisfied.Condition 1: ( k_1 < m K_B (1 - frac{d}{k_2}) )If this holds, then ( lambda_1 < 0 ), and since ( lambda_2 < 0 ), the equilibrium is stable.If ( lambda_1 > 0 ), then it's unstable.3. Equilibrium ( (P_A^*, 0) ) where ( P_A^* = K_A ):Evaluate ( J ) at ( (K_A, 0) ):First row, first column: ( k_1 (1 - frac{2 K_A}{K_A}) - m cdot 0 = k_1 (1 - 2) = -k_1 )First row, second column: ( -m K_A )Second row, first column: ( m cdot 0 = 0 )Second row, second column: ( k_2 (1 - 0) + m K_A - d = k_2 + m K_A - d )So, Jacobian:[ J(K_A, 0) = begin{bmatrix}- k_1 & -m K_A 0 & k_2 + m K_A - dend{bmatrix} ]Eigenvalues:- ( lambda_1 = -k_1 ) (negative)- ( lambda_2 = k_2 + m K_A - d )For stability, ( lambda_2 < 0 ):( k_2 + m K_A - d < 0 ) → ( m K_A < d - k_2 )But since ( k_2 ) is a growth rate, it's positive, and ( d ) is a death rate, also positive. However, ( m K_A ) is positive, so ( d - k_2 ) must be positive for this to hold, i.e., ( d > k_2 ). But in this equilibrium, ( P_A = K_A ), which is the carrying capacity for region A, and region B is zero. If ( d > k_2 ), region B's death rate dominates its growth, which might make sense.However, the equilibrium ( (P_A^*, 0) ) exists regardless of ( k_2 ) and ( d ), but its stability depends on whether ( k_2 + m K_A - d < 0 ).4. Non-trivial equilibrium ( (P_A^*, P_B^*) ):This is the most complex case. We need to evaluate the Jacobian at ( (P_A^*, P_B^*) ) and find its eigenvalues.From earlier, we have:( P_B^* = frac{k_1}{m} (1 - frac{P_A^*}{K_A}) )And ( P_A^* ) is given by:( P_A^* = frac{k_2 k_1 - (k_2 - d) m K_B}{frac{k_2 k_1}{K_A} + m^2 K_B} )This expression is quite involved, so instead of computing the Jacobian directly, perhaps we can analyze the stability conditions based on the trace and determinant.The eigenvalues of the Jacobian ( J ) are given by:( lambda = frac{1}{2} left[ Tr(J) pm sqrt{(Tr(J))^2 - 4 Det(J)} right] )Where ( Tr(J) ) is the trace and ( Det(J) ) is the determinant.For stability, we need both eigenvalues to have negative real parts. This occurs if:1. ( Tr(J) < 0 )2. ( Det(J) > 0 )So, let's compute ( Tr(J) ) and ( Det(J) ) at ( (P_A^*, P_B^*) ).From the Jacobian:( Tr(J) = [k_1 (1 - frac{2 P_A^*}{K_A}) - m P_B^*] + [k_2 (1 - frac{2 P_B^*}{K_B}) + m P_A^* - d] )Simplify:( Tr(J) = k_1 (1 - frac{2 P_A^*}{K_A}) - m P_B^* + k_2 (1 - frac{2 P_B^*}{K_B}) + m P_A^* - d )From the equilibrium conditions, we have:From equation 1: ( k_1 (1 - frac{P_A^*}{K_A}) = m P_B^* )From equation 2: ( k_2 (1 - frac{P_B^*}{K_B}) + m P_A^* = d )Let me use these to simplify ( Tr(J) ):First, note that:( k_1 (1 - frac{2 P_A^*}{K_A}) = k_1 (1 - frac{P_A^*}{K_A}) - k_1 frac{P_A^*}{K_A} = m P_B^* - k_1 frac{P_A^*}{K_A} )Similarly, ( k_2 (1 - frac{2 P_B^*}{K_B}) = k_2 (1 - frac{P_B^*}{K_B}) - k_2 frac{P_B^*}{K_B} = (d - m P_A^*) - k_2 frac{P_B^*}{K_B} ) (from equation 2)Substitute these into ( Tr(J) ):( Tr(J) = [m P_B^* - k_1 frac{P_A^*}{K_A}] - m P_B^* + [(d - m P_A^*) - k_2 frac{P_B^*}{K_B}] + m P_A^* - d )Simplify term by term:1. ( m P_B^* - k_1 frac{P_A^*}{K_A} )2. ( - m P_B^* )3. ( d - m P_A^* - k_2 frac{P_B^*}{K_B} )4. ( + m P_A^* )5. ( - d )Combine like terms:- ( m P_B^* - m P_B^* = 0 )- ( - k_1 frac{P_A^*}{K_A} )- ( d - d = 0 )- ( - m P_A^* + m P_A^* = 0 )- ( - k_2 frac{P_B^*}{K_B} )Thus, ( Tr(J) = - k_1 frac{P_A^*}{K_A} - k_2 frac{P_B^*}{K_B} )Since ( k_1, k_2, P_A^*, P_B^*, K_A, K_B ) are all positive, ( Tr(J) < 0 ).Now, compute ( Det(J) ):( Det(J) = [k_1 (1 - frac{2 P_A^*}{K_A}) - m P_B^*][k_2 (1 - frac{2 P_B^*}{K_B}) + m P_A^* - d] - (-m P_A^*)(m P_B^*) )Simplify:( Det(J) = [k_1 (1 - frac{2 P_A^*}{K_A}) - m P_B^*][k_2 (1 - frac{2 P_B^*}{K_B}) + m P_A^* - d] + m^2 P_A^* P_B^* )Again, using the equilibrium conditions:From equation 1: ( k_1 (1 - frac{P_A^*}{K_A}) = m P_B^* ) → ( k_1 (1 - frac{2 P_A^*}{K_A}) = m P_B^* - k_1 frac{P_A^*}{K_A} )From equation 2: ( k_2 (1 - frac{P_B^*}{K_B}) + m P_A^* = d ) → ( k_2 (1 - frac{2 P_B^*}{K_B}) = d - m P_A^* - k_2 frac{P_B^*}{K_B} )Substitute these into ( Det(J) ):( Det(J) = [m P_B^* - k_1 frac{P_A^*}{K_A}][d - m P_A^* - k_2 frac{P_B^*}{K_B} + m P_A^* - d] + m^2 P_A^* P_B^* )Simplify the second bracket:( [d - m P_A^* - k_2 frac{P_B^*}{K_B} + m P_A^* - d] = - k_2 frac{P_B^*}{K_B} )Thus,( Det(J) = [m P_B^* - k_1 frac{P_A^*}{K_A}] (- k_2 frac{P_B^*}{K_B}) + m^2 P_A^* P_B^* )Expand:( Det(J) = - m P_B^* k_2 frac{P_B^*}{K_B} + k_1 frac{P_A^*}{K_A} k_2 frac{P_B^*}{K_B} + m^2 P_A^* P_B^* )Factor out ( P_B^* ):( Det(J) = P_B^* [ - m k_2 frac{P_B^*}{K_B} + k_1 k_2 frac{P_A^*}{K_A K_B} + m^2 P_A^* ] )This expression is quite complex, but we can analyze its sign.Given that all parameters are positive, the sign of ( Det(J) ) depends on the terms inside the brackets.However, without specific values, it's challenging to determine the sign definitively. But generally, for the non-trivial equilibrium to be stable, we need ( Det(J) > 0 ) and ( Tr(J) < 0 ). Since we've already established ( Tr(J) < 0 ), the stability hinges on ( Det(J) > 0 ).Given the complexity, perhaps we can consider specific conditions under which ( Det(J) > 0 ). For instance, if the positive terms outweigh the negative ones.Alternatively, perhaps we can use the expressions for ( P_A^* ) and ( P_B^* ) to substitute back into ( Det(J) ) and see if it simplifies.Given the time constraints, I think it's reasonable to conclude that the non-trivial equilibrium ( (P_A^*, P_B^*) ) is stable if the determinant is positive, which would require certain relationships between the parameters. Specifically, the interaction terms and growth/death rates must balance in a way that the positive feedbacks (like growth) are outweighed by negative feedbacks (like competition and death).In summary, the equilibrium points are:1. Trivial: ( (0, 0) ) - unstable2. ( (0, P_B^*) ) - stable if ( k_1 < m K_B (1 - frac{d}{k_2}) ) and ( k_2 > d )3. ( (P_A^*, 0) ) - stable if ( k_2 + m K_A < d )4. Non-trivial ( (P_A^*, P_B^*) ) - stable if ( Det(J) > 0 ), which depends on parameter relationshipsFor both regions to achieve stable coexistence, the non-trivial equilibrium must be stable. This requires that the determinant of the Jacobian at that point is positive, which translates to certain inequalities among the parameters. Specifically, the combined effects of growth, migration, and death rates must allow for a balance where neither population outcompetes the other, leading to a stable coexistence.Therefore, the conditions for stable coexistence are:- ( k_2 > d ) (so that region B can sustain a positive population)- ( k_1 < m K_B (1 - frac{d}{k_2}) ) (to ensure region A doesn't grow too much)- Additionally, the determinant condition must hold, which is more complex but generally implies that the interaction terms and growth rates are balanced appropriately.In terms of the nature of the equilibrium points, the trivial equilibrium is a saddle point, the axis equilibria are nodes (stable or unstable depending on parameters), and the non-trivial equilibrium can be a stable node or spiral depending on the eigenvalues' nature (real or complex).</think>

answer:<think>Okay, so I'm trying to solve this problem about the Mi'kmaq cultural gathering. It involves some geometry, specifically circles and areas. Let me break it down step by step.First, part 1: The elder wants 5 concentric circles around the central performance area. The innermost circle has a radius of 5 meters, and each subsequent circle increases by 3 meters. I need to calculate the total area available for seating.Hmm, concentric circles mean they all share the same center. So, the seating area is the area between the innermost circle and the outermost circle. Wait, actually, since there are 5 concentric circles, does that mean there are 4 annular regions for seating? Because the innermost is the performance area, and then each circle after that adds a ring for seating.Let me visualize this. The central performance area is a circle with radius 5m. Then, the first seating ring would be from 5m to 8m (5+3), the second from 8m to 11m, the third from 11m to 14m, and the fourth from 14m to 17m. So, actually, there are 4 seating rings, each 3 meters wide, around the central area.Wait, but the problem says 5 concentric circles, so maybe the performance area is the first circle, and then 4 more circles for seating? Or is the performance area the innermost, and then 5 seating circles? Hmm, the wording says "the seating area will be arranged in 5 concentric circles around the central performance area." So, the central performance area is one circle, and then 5 concentric circles around it for seating. So, that would make 6 circles in total? Wait, no, maybe not.Wait, let me read again: "the seating area will be arranged in 5 concentric circles around the central performance area." So, the central performance area is one circle, and then 5 concentric circles around it for seating. So, the total number of circles is 6: central performance area, then 5 seating circles. But each subsequent circle's radius increases by 3 meters. So, starting from 5m, each next circle is 3m more.Wait, but if the central performance area is 5m radius, then the first seating circle would be 5+3=8m, the second 8+3=11m, third 14m, fourth 17m, fifth 20m. So, the outermost circle has a radius of 20m.But the seating area is the area between the central performance area and the outermost circle. So, the total seating area would be the area of the outermost circle minus the area of the central performance area.Wait, but the problem says "the seating area will be arranged in 5 concentric circles." So, does that mean that each seating ring is a separate circle? Or is it just that the seating is divided into 5 concentric circles? Hmm, maybe I need to think differently.Alternatively, perhaps the seating area is divided into 5 concentric rings, each 3 meters wide. So, starting from 5m, the first ring is 5m to 8m, the second 8m to 11m, and so on until the fifth ring. Wait, but 5 rings would mean the outermost radius is 5 + 5*3 = 20m. So, the total seating area is the area from 5m to 20m.But the problem says "the seating area will be arranged in 5 concentric circles around the central performance area." So, maybe each seating area is a circle, not an annulus. Hmm, that doesn't quite make sense because if you have 5 concentric circles, each with increasing radii, the area between each circle would be the seating areas.Wait, perhaps the total seating area is the area of the outermost circle minus the central performance area. So, if there are 5 concentric circles, each 3m apart, starting from 5m, then the radii would be 5, 8, 11, 14, 17, 20 meters. So, the outermost circle is 20m. Therefore, the total seating area is the area of the circle with radius 20m minus the area of the central performance area (radius 5m).But wait, the problem says "the seating area will be arranged in 5 concentric circles." So, maybe each of those 5 circles is a seating area, but that would imply 5 separate circles, which doesn't make much sense. Alternatively, perhaps the seating area is divided into 5 concentric rings, each 3 meters wide, so 5 rings in total.Wait, let me think again. The central performance area is a circle with radius 5m. Then, around it, there are 5 concentric circles for seating. Each subsequent circle's radius increases by 3 meters. So, starting from 5m, the first seating circle is 8m, then 11m, 14m, 17m, and 20m. So, the seating area is the area between 5m and 20m. Therefore, the total seating area is the area of the circle with radius 20m minus the area of the circle with radius 5m.Yes, that makes sense. So, the total seating area is π*(20)^2 - π*(5)^2 = π*(400 - 25) = π*375 ≈ 1178.097 square meters.Wait, but the problem says "the seating area will be arranged in 5 concentric circles." So, does that mean that each of those 5 circles is a separate seating area, or is it the area between the central circle and the outermost circle? I think it's the latter because otherwise, if each circle is a seating area, they would overlap, which doesn't make sense.So, I think the total seating area is the area from 5m to 20m, which is π*(20^2 - 5^2) = π*(400 - 25) = 375π square meters.But let me double-check. If there are 5 concentric circles, starting from 5m, each 3m apart, then the radii are 5, 8, 11, 14, 17, 20. So, the outermost circle is 20m. Therefore, the seating area is the area from 5m to 20m, which is the area of the 20m circle minus the 5m circle.Yes, that seems correct.Now, moving on to part 2: The elder wants to divide the circular seating area into 8 equal radial sections for different families. Each section extends from the innermost circle to the outermost circle. I need to calculate the area of each radial section and the length of the arc that forms the boundary of one section at the outermost circle.So, the seating area is an annulus with inner radius 5m and outer radius 20m. Dividing this into 8 equal radial sections means each section is like a slice of a pie, with two straight edges (radii) and an arc.The area of each radial section would be the area of the annulus divided by 8. The area of the annulus is π*(20^2 - 5^2) = 375π, as calculated earlier. So, each section's area is 375π / 8 ≈ 46.875π square meters.Alternatively, since each section is a sector of the outer circle minus a sector of the inner circle, both with the same central angle. Since there are 8 equal sections, each central angle is 360/8 = 45 degrees, or π/4 radians.So, the area of each radial section can also be calculated as (1/2)*r^2*θ - (1/2)*r_inner^2*θ, where θ is π/4 radians.Plugging in the numbers: (1/2)*(20^2)*(π/4) - (1/2)*(5^2)*(π/4) = (1/2)*(400)*(π/4) - (1/2)*(25)*(π/4) = (200)*(π/4) - (12.5)*(π/4) = (200 - 12.5)*(π/4) = 187.5*(π/4) = 46.875π square meters. So, same result.Now, the length of the arc at the outermost circle for each section. The circumference of the outermost circle is 2π*20 = 40π meters. Since it's divided into 8 equal sections, each arc length is 40π / 8 = 5π meters.Alternatively, the arc length is given by θ*r, where θ is in radians. θ is π/4, so arc length is (π/4)*20 = 5π meters. Same result.So, summarizing:1. Total seating area: 375π square meters.2. Each radial section area: 46.875π square meters, and arc length: 5π meters.Wait, but let me make sure about part 1 again. The problem says "the seating area will be arranged in 5 concentric circles around the central performance area." So, does that mean that each of those 5 circles is a separate seating area, or is it the area between the central circle and the outermost circle?If it's 5 concentric circles, each 3m apart, starting from 5m, then the outermost circle is 5 + 5*3 = 20m. So, the seating area is the area between 5m and 20m, which is 375π.Alternatively, if each of the 5 circles is a seating area, meaning 5 separate circles, but that would mean overlapping areas, which doesn't make sense. So, I think the first interpretation is correct.Therefore, the total seating area is 375π square meters.For part 2, each radial section is 1/8 of the annulus, so area is 375π /8, and the arc length is 5π meters.I think that's it.</think>

answer:<think>Alright, so I've got this problem about designing a scoring system for a strategy game. It's broken down into two parts, both involving some graph theory concepts. Let me try to wrap my head around each part step by step.Starting with the first part: Graph Construction and Path Optimization. The game environment is represented as a directed graph where nodes are game states and edges are actions with associated scores. I need to find the path from node S to node T that maximizes the total score. But wait, there's a twist—some nodes have penalties that reduce the total score if visited. Hmm, okay, so it's not just about the edges but also the nodes themselves affecting the score.So, how do I model this? I remember that in graph theory, when dealing with paths and maximizing scores, it's often related to the longest path problem. But in this case, it's a bit more complex because nodes can have penalties. Let me think about how to incorporate both edge weights and node penalties into the optimization.First, let's define the problem mathematically. The graph G has nodes V and edges E. Each edge (u, v) has a weight w(u, v), which is the score for taking that action. Additionally, each node v has a penalty p(v) that subtracts from the total score if the node is visited. So, the total score for a path is the sum of the edge weights minus the sum of penalties for all nodes visited along the path, except maybe the starting node? Or does the starting node also have a penalty?Wait, the problem says "if visited," so I think the starting node S might not have a penalty because the game starts there, but I'm not entirely sure. Maybe I should clarify that in my formulation. For now, I'll assume that all nodes except S have penalties if visited.So, the total score for a path P from S to T would be the sum of w(u, v) for each edge (u, v) in P minus the sum of p(v) for each node v in P, excluding S. Or maybe including S? Hmm, the problem statement isn't clear on whether the starting node incurs a penalty. It just says "if visited," so perhaps S does have a penalty if it's considered visited. But since the game starts there, maybe it's not penalized. I'll need to make an assumption here.Let me proceed with the assumption that the starting node S does not have a penalty. So, the total score is the sum of all edge weights along the path minus the sum of penalties for all nodes visited after S.Now, how do I model this as an optimization problem? It seems like a variation of the longest path problem where each node adds a penalty, which is a negative weight. So, perhaps I can transform the graph into one where each node's penalty is incorporated into the edges.Wait, another approach is to adjust the edge weights to account for the penalties. If moving into a node v incurs a penalty p(v), then perhaps each edge leading into v should have its weight reduced by p(v). But that might not capture it correctly because the penalty is applied once per visit, not per edge.Alternatively, maybe I can model this by adding a self-loop on each node v with a weight of -p(v). But that might complicate things because it could create cycles, and the problem is about finding a path from S to T, which is acyclic.Hmm, perhaps a better way is to adjust the node values. Let me think in terms of dynamic programming. For each node v, the maximum score to reach v is the maximum over all incoming edges (u, v) of (max score to reach u) + w(u, v) - p(v). But wait, that would subtract the penalty p(v) each time we reach v, which might not be correct if we visit v multiple times. But in a simple path, we don't visit nodes multiple times, so maybe that's acceptable.Wait, no, in a path from S to T, each node is visited at most once, so the penalty p(v) is subtracted only once if v is on the path. So, perhaps the total score is the sum of edge weights minus the sum of penalties for all nodes except S on the path.Therefore, the optimization problem can be formulated as finding a path P from S to T such that the total score is maximized, where the total score is:Total = sum_{(u, v) ∈ P} w(u, v) - sum_{v ∈ P, v ≠ S} p(v)So, mathematically, we can model this as a graph where each node v has a value of -p(v), and we need to find the path from S to T that maximizes the sum of edge weights plus the sum of node values (which are negative). But in standard graph problems, nodes don't have weights, only edges. So, how do we incorporate node penalties?One way is to split each node v into two nodes: v_in and v_out. Then, for each edge (u, v), we connect u_out to v_in with weight w(u, v). Then, we add an edge from v_in to v_out with weight -p(v). This way, every time we enter a node v, we have to pay the penalty p(v) to exit it. This transformation allows us to model the node penalties as edge weights.So, the transformed graph G' would have nodes V' = {v_in, v_out | v ∈ V}. The edges in G' would be:- For each edge (u, v) in E, add an edge from u_out to v_in with weight w(u, v).- For each node v in V, add an edge from v_in to v_out with weight -p(v).Then, the problem reduces to finding the longest path from S_out to T_in in G', because S starts at S_out (since we don't penalize S) and T is the end, so we don't need to exit T.Wait, but in the transformed graph, we need to start at S_out and end at T_in? Or maybe S_in and T_out? Let me think.If S is the starting node, in the original graph, we don't have a penalty for S. So in G', we should start at S_out because S_in would have an edge to S_out with weight -p(S), but we don't want to penalize S. Therefore, the starting point is S_out, and the ending point is T_in because once we reach T, we don't need to exit it (and thus don't incur any penalty for T, assuming T doesn't have a penalty or we don't exit it).Wait, but in the problem statement, it's not specified whether T has a penalty. It just says "if visited." So if T is visited, we have to subtract p(T). But in the path, T is the end, so we have to include it. So, in the transformed graph, we need to include the penalty for T as well.Hmm, maybe I should adjust the starting and ending points. Let me try again.In G', each node v is split into v_in and v_out. For each edge (u, v) in E, we have u_out -> v_in with weight w(u, v). For each node v, we have v_in -> v_out with weight -p(v). Now, to model the starting point S, since we don't want to penalize S, we can start at S_out. Similarly, for T, since we do want to penalize it if it's visited, we need to end at T_out, but wait, in the path, we have to end at T, so perhaps we should end at T_in? No, because the path ends at T, so we don't exit it, hence we don't have to pay the penalty for T.Wait, this is getting confusing. Let me think differently.Suppose we model the penalty for each node v as a cost that is incurred when entering v. So, when you traverse an edge into v, you pay the penalty p(v). But in the problem, the penalty is subtracted if the node is visited, regardless of how you got there. So, each time you visit v, you subtract p(v). But in a simple path, you visit each node at most once, so it's just a one-time subtraction.Therefore, the total score is the sum of all edge weights on the path minus the sum of penalties for all nodes on the path except S.So, to model this, we can adjust the edge weights to include the penalty of the destination node. That is, for each edge (u, v), we can redefine its weight as w(u, v) - p(v). Then, the problem becomes finding the path from S to T with the maximum sum of these adjusted weights.But wait, this would subtract p(v) for each edge entering v, which would effectively subtract p(v) for each time we enter v. But in a simple path, we enter each node at most once, so this would correctly subtract p(v) once for each node v on the path, except S.Wait, but S is the starting node, so it's included in the path, but we don't subtract p(S). So, if we adjust all edges to subtract p(v) for their destination, then when we start at S, the first edge from S to some u would subtract p(u), which is correct. But S itself isn't subtracted because it's the starting point. So, this seems to handle it correctly.Therefore, the mathematical formulation can be:Maximize the sum over edges (u, v) in the path P of [w(u, v) - p(v)], where P is a path from S to T.But we need to ensure that S is included in the path without subtracting p(S). Since S is the starting node, it's part of the path, but we don't subtract p(S) because we don't have an edge leading into S (except maybe a self-loop, but that's not part of the path). So, this approach works because the penalty for S is not subtracted since there's no edge leading into S in the path.Therefore, the optimization problem is to find a path P from S to T such that the sum of [w(u, v) - p(v)] for all edges (u, v) in P is maximized.Alternatively, we can model this as a standard longest path problem where each edge (u, v) has weight w(u, v) - p(v), and we need to find the longest path from S to T in this modified graph.But wait, in the standard longest path problem, we don't have node penalties, so this modification effectively incorporates the node penalties into the edge weights. That makes sense.So, to summarize, the mathematical formulation is:Maximize Σ [w(u, v) - p(v)] for all edges (u, v) in P, where P is a path from S to T.Now, moving on to the second part: Probability and Expected Value Calculation. Each action (edge) has a success probability P(u, v). If successful, the player gets w(u, v); if not, zero. We need to compute the expected value of the total score for the optimal path from S to T, considering the probabilities.This seems like a problem involving expected values along a path where each edge's contribution is probabilistic. So, for each edge in the path, the expected score is P(u, v) * w(u, v). The total expected score is the sum of these expected values for all edges in the path.But wait, the path is optimal in terms of maximizing the expected value. So, we need to find the path from S to T that maximizes the expected total score, considering the probabilities of each edge's success.This is similar to finding the longest path in a graph where each edge has a weight of P(u, v) * w(u, v). So, the expected value for each edge is known, and we need to find the path with the maximum sum of these expected values.Therefore, the expected value of the total score for the optimal path is the sum over all edges (u, v) in the optimal path P of [P(u, v) * w(u, v)].But wait, in the first part, we had to consider node penalties. Does that affect the expected value calculation? Yes, because the penalties are subtracted from the total score. So, in the expected value, we also need to account for the penalties.Wait, the penalties are deterministic once a node is visited. So, if a node v is visited, the penalty p(v) is subtracted regardless of the success of the edges. So, the penalty is a sure loss if the node is visited.Therefore, the expected total score is the expected sum of the edge scores minus the sum of penalties for all nodes visited.So, the expected value E is:E = [Sum over edges (u, v) in P of P(u, v) * w(u, v)] - [Sum over nodes v in P, v ≠ S of p(v)]But wait, the penalties are only subtracted if the node is visited, which is certain if the path P is taken. So, the penalties are deterministic, and the edge scores are probabilistic.Therefore, the expected total score is the sum of the expected scores from each edge in P minus the sum of penalties for each node in P (except S).So, the expected value is:E = Σ [P(u, v) * w(u, v)] - Σ [p(v)]where the first sum is over all edges (u, v) in P, and the second sum is over all nodes v in P, excluding S.Therefore, to find the optimal path, we need to maximize E, which is the sum of expected edge scores minus the sum of node penalties.This can be modeled as finding the path from S to T that maximizes:Σ [P(u, v) * w(u, v)] - Σ [p(v)]for all edges (u, v) in P and nodes v in P (excluding S).Alternatively, we can adjust the edge weights to include the expected value and then subtract the node penalties. So, for each edge (u, v), redefine its weight as P(u, v) * w(u, v). Then, the problem becomes finding the longest path from S to T in this modified graph, but also subtracting the sum of penalties for all nodes on the path (excluding S).But how do we incorporate both the expected edge weights and the node penalties into a single optimization problem?One approach is to model this as a graph where each edge (u, v) has a weight of P(u, v) * w(u, v) - p(v). But wait, that would subtract p(v) for each edge, which might not be correct because p(v) is subtracted once per node, not per edge.Alternatively, we can split each node into two as before, but now the edge weights are the expected values. So, for each edge (u, v), we have u_out -> v_in with weight P(u, v) * w(u, v). Then, for each node v, we have v_in -> v_out with weight -p(v). Then, the problem becomes finding the longest path from S_out to T_in in this transformed graph.Wait, but in this case, the penalty is subtracted when exiting the node, which corresponds to visiting it. So, if we start at S_out, we don't subtract p(S), and when we exit any other node v, we subtract p(v). This seems correct.Therefore, the expected value can be computed by finding the longest path in this transformed graph, where edges have weights P(u, v) * w(u, v) and node penalties are modeled as edges from v_in to v_out with weight -p(v).So, the expected value E is the maximum total weight of a path from S_out to T_in in this transformed graph.Alternatively, without transforming the graph, we can think of it as:E = max_P [Σ (P(u, v) * w(u, v)) - Σ p(v)] where P is a path from S to T, and the second sum is over all nodes v in P except S.Therefore, the expected value is the maximum over all possible paths P of the sum of expected edge scores minus the sum of penalties for nodes visited.So, to compute this, we can use dynamic programming where for each node v, we keep track of the maximum expected value to reach v, considering the penalties.Let me try to formalize this.Let’s define dp[v] as the maximum expected value to reach node v from S, considering the penalties for all nodes along the path.Then, for each node v, dp[v] = max over all predecessors u of [dp[u] + P(u, v) * w(u, v) - p(v)].Wait, no. Because p(v) is subtracted when we visit v, which is after arriving at v. So, when we arrive at v via edge (u, v), we add the expected value of that edge and then subtract p(v) because we've visited v.But in the DP formulation, we need to ensure that p(v) is subtracted only once when we first visit v. However, in a path, each node is visited at most once, so the DP can be structured as:dp[v] = max over all u such that (u, v) ∈ E of [dp[u] + P(u, v) * w(u, v) - p(v)]But wait, this would subtract p(v) for every edge leading into v, which is incorrect because p(v) should be subtracted only once when v is first visited. However, in a simple path, each node is visited once, so the subtraction is correct.Wait, no. If we use this DP formulation, for each edge (u, v), we are considering the expected value of taking that edge and then subtracting p(v). But if we have multiple edges leading into v, each would subtract p(v), which would incorrectly subtract p(v) multiple times if we consider multiple paths to v.But in reality, once we've subtracted p(v) when we first reach v, we shouldn't subtract it again if we consider another path to v. However, in a simple path, we don't revisit nodes, so each node is reached once, and thus p(v) is subtracted once.Wait, but in the DP formulation, dp[v] represents the maximum expected value to reach v, considering all possible paths to v. So, if we have multiple paths to v, each path would have subtracted p(v) once. Therefore, the DP formulation as:dp[v] = max over u [dp[u] + P(u, v) * w(u, v) - p(v)]is correct because for each path to v, we subtract p(v) once.But wait, actually, when we reach v via edge (u, v), we add the expected value of that edge and then subtract p(v). So, the DP transition is:dp[v] = max over u [dp[u] + P(u, v) * w(u, v)] - p(v)Wait, no, because p(v) is a penalty for visiting v, which is a one-time cost when you first visit v. So, it should be subtracted once when you reach v, regardless of how you got there.Therefore, the correct DP transition is:dp[v] = max over u [dp[u] + P(u, v) * w(u, v)] - p(v)This way, for each node v, we first compute the maximum expected value to reach any predecessor u, add the expected value of the edge (u, v), and then subtract p(v) because we've now visited v.This makes sense because the penalty is incurred upon visiting v, which happens after traversing the edge into v.Therefore, the DP formulation is:dp[v] = max_{u ∈ predecessors(v)} [dp[u] + P(u, v) * w(u, v)] - p(v)with the base case dp[S] = 0 (since we start at S without any penalty).Then, the expected value of the optimal path is dp[T].So, putting it all together, the expected value E is:E = dp[T] = max_{P} [Σ (P(u, v) * w(u, v)) - Σ p(v)]where the first sum is over edges in P and the second sum is over nodes in P except S.Therefore, the mathematical formulation for the expected value is the maximum over all paths P from S to T of the sum of expected edge scores minus the sum of penalties for nodes visited.In summary, for part 1, the optimization problem is to find the path from S to T that maximizes the sum of edge weights minus node penalties. For part 2, it's to find the path that maximizes the expected sum of edge scores minus node penalties, considering the success probabilities of each edge.I think I've got a handle on both parts now. The key was to model the node penalties appropriately, either by transforming the graph or adjusting the DP formulation to account for them.</think>